6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm Hg. Find it's molar mass (show your work)

Answer:

0.82V

Explanation:

0.48V+0.34V=0.82V

Answer:

-.14

Explanation:

Answer:

[tex]y_{O2} =4.3[/tex]%

Explanation:

The ethanol combustion reaction is:

[tex]C_{2}H_{5} OH+3O_{2}[/tex]→[tex]2CO_{2}+3H_{2}O[/tex]

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

[tex]x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}[/tex]

Dividing the previous equation by x:

[tex]1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}[/tex]

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

[tex]3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )[/tex]

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

[tex]3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles[/tex]

Calculate the number of moles of CO2 and water considering the same:

[tex]0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)[/tex]

[tex]0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)[/tex]

The total number of moles at the reactor output would be:

[tex]N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)[/tex]

So, the oxygen mole fraction would be:

[tex]y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3[/tex]%

Answer:

[tex]v_{in} =6.373 \frac{m}{s}[/tex]

[tex]v_{out} =2.832 \frac{m}{s}[/tex]

[tex]W=0.63063*0.5=0.3153kW[/tex]

Explanation:

First, we are going to need the water specific volume at 15ºC: v=0.001001 [tex]\frac{m^{3}}{kg}[/tex]. The density '[tex]p[/tex]' of the water is the inverse of the specific volume: [tex]p=999 \frac{kg}{m^{3} }[/tex]

First, consider the mass flow, which is related to the volumetric flow (density and velocity) and the area:

[tex]m=pvA[/tex]

The area of each cross-section is:

[tex]A_{in} =\frac{\pi*0.01^{2} }{4}=7.854*10^{-5}[/tex] (in square meters). Here, the radius was not used but the diameter, which means a division by 4 (2 squared).

[tex]A_{out} =\frac{\pi*0.015^{2} }{4}=1.767*10^{-4}m^{2}[/tex]

From mass flow isolate the velocity and calculate it:

[tex]v=\frac{m}{pA}[/tex]

[tex]v_{in} =\frac{0.5}{999*7.854*10^{-5} }=6.373 \frac{m}{s}[/tex]

[tex]v_{out} =\frac{0.5}{999*1.767*10^{-4} }=2.832 \frac{m}{s}[/tex]

The work of the pump is calculated considering an energy balance on the pump:

[tex]w=h_{in}-h_{out}[/tex]

Considering the isentropic process may give us the relation:

[tex]dh=vdP\\ h_{2} -h_{1}=v*(P_{2} -P_{1})[/tex]

Applying that to the pump,

[tex]w=-0.001001*(700-70)=-0.63063 \frac{kJ}{kg}[/tex]

Multiplying it by the mass flow:

[tex]W=-0.63063*0.5=-0.3153kW[/tex]

The work is negative because it is entering to the system, but the required is positive. (It is just a standard rule)

The question involves applying the continuity equation in physics to calculate the velocities of water at the inlet and outlet of a pump, and computing the work done by the pump in increasing water pressure.

To answer the question about the velocity of the water at the inlet and outlet of the pump, we can apply the principle of continuity equation which states that the density times the area times the velocity must remain constant. Since the density of the water does not change, the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region.

The work done by the pump can be calculated, knowing that it can be used to increase the kinetic energy and gain potential energy as the pressure increases.

The power required to operate the pump can be computed from these parameters. A substantial amount of power requires a large pump, such as the one in the question which increases the water's pressure from 70 kPa to 700 kPa. This drastic increase in pressure indicates that a significant amount of work is being done on the fluid.

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Answer:

TRUE

Explanation:

Isozymes are enzymes that differ in amino acid sequence but catalyze the same chemical reaction.  

These enzymes usually have different enzyme kinetics, that is, they have different Km and Vmax values,

For example, three isozymes of lactate dehydrogenase (A, B, and C) have Km values of 0.260, 0.172, and 0.052 mmol/L, respectively.

Answer: The volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.

Explanation:

[tex]PV=nRT[/tex]

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g    (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]

T = temperature of container = [tex]60^oC=[60+273]K=333K[/tex]

Putting values in above equation, we get:

[tex]700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L[/tex]

Converting this value into [tex]m^3[/tex], we use the conversion factor:

[tex]1m^3=1000L[/tex]

So, [tex]\Rightarrow (\frac{1m^3}{1000L})\times 154.21L[/tex]

[tex]\Rightarrow 0.1542m^3[/tex]

[tex]U=\frac{3}{2}nRT[/tex]

or,

[tex]U=\frac{3}{2}\frac{m}{M}RT[/tex]

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g  (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = [tex]8.314J/K.mol[/tex]

T = temperature = [tex]60^oC=[60+273]K=333K[/tex]

Putting values in above equation, we get:

[tex]U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J[/tex]

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.

You start with 100 kg raw potatoes.

8% of mass is lost by peeling:

100 - ( 8 / 100) × 100 = 92 kg of peeled raw potatoes

By drying the water contents is reduces from 75% to 7%, that means a 68% loss by weight:

92 - ( 68 / 100) × 92 = 92 - 62.56 = 29.44 kg of dried potatoes

The mass of dried potatoes produced from each 100 kg of raw potatoes is 24.7312 kg.

To solve this problem, we will follow the steps outlined in the question:

1. Start with 100 kg of raw potatoes.

2. Calculate the loss of mass due to peeling. Since 8% by weight is lost, we have:

 Mass lost due to peeling = 8% of 100 kg = 0.08 * 100 kg = 8 kg.

3. The remaining mass after peeling is:

 Mass after peeling = Initial mass - Mass lost due to peeling = 100 kg - 8 kg = 92 kg.

4. Now, we need to adjust the water content from 75% to 7%. Let's denote the mass of the dried potatoes as m. The amount of water to be removed from the potatoes is the difference between the initial water content and the final water content:

 Water to be removed = (75% - 7%) * m.

5. Since the initial water content is 75%, the amount of water in the potatoes after peeling is:

 Initial water mass = 75% of 92 kg = 0.75 * 92 kg = 69 kg.

6. The final water content should be 7%, so the mass of water in the dried potatoes will be:

 Final water mass = 7% of m = 0.07 * m.

7. The difference in water content is the amount of water that needs to be removed:

 Water to be removed = Initial water mass - Final water mass.

8. Since the mass of the dried potatoes \( m \) consists of 7% water and 93% solids (including starch), we can express the mass of the solids as:

Solids mass = 93% of m = 0.93 * m.

9. The solids mass after drying must be equal to the solids mass after peeling, which is the total mass after peeling minus the initial water mass:

 Solids mass after peeling = Mass after peeling - Initial water mass = 92 kg - 69 kg = 23 kg.

10. Now we can set up the equation:

Solids mass after drying = Solids mass after peeling,

0.93 * m = 23 kg.

11. Solving for m, we get:

[tex]\( m = \frac{23 \text{ kg}}{0.93} \).[/tex]

12. Calculate the mass of the dried potatoes:

[tex]\( m = \frac{23}{0.93} \approx 24.7312 \) kg.[/tex]

Therefore, the mass of dried potatoes produced from each 100 kg of raw potatoes is approximately 24.7312 kg.

The answer is: 24.7312.

Answer: The molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

For lead chloride:

Given mass of lead chloride = 18.86 g

Molar mass of lead chloride = 278.1 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of lead chloride}=\frac{18.86g}{278.1g/mol}=0.068mol[/tex]

[tex]Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3[/tex]

By Stoichiometry of the reaction:

1 mole of lead chloride is formed by 1 mole of lead nitrate

So, 0.068 moles of lead chloride will be formed from = [tex]\frac{1}{1}\times 0.068=0.068mol[/tex] of lead nitrate

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Volume of solution = 200 mL = 0.200 L   (Conversion factor: 1 L = 1000 mL)

Moles of lead nitrate = 0.068 moles

Putting values in above equation, we get:

[tex]\text{Molarity of }Pb(NO_3)_2\text{ solution}=\frac{0.068mol}{0.065L}\\\\\text{Molarity of }Pb(NO_3)_2\text{ solution}=0.34M[/tex]

Hence, the molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.

The molarity of the Pb(NO3)2(aq) solution can be calculated by first finding the number of moles of PbCl2 formed in the reaction and then dividing it by the original volume of Pb(NO3)2 solution. The calculated molarity of the Pb(NO3)2(aq) solution is 0.339 M.

The calculation of molarity of the Pb(NO3)2(aq) solution is necessary in this case. In this precipitation reaction, lead nitrate reacts with sodium chloride to form lead chloride, which precipitates out, and sodium nitrate. The molar mass of lead chloride (PbCl2) is 278.1 g/mol.

Step 1: Find the number of moles of PbCl2. For this, you can divide the mass of the precipitate (PbCl2) obtained by the molar mass of PbCl2. So, the number of moles are 18.86 g / 278.1 g/mol = 0.0678 mol.

Step 2: Calculate the amount in liters of original solution of lead nitrate. Here, since the given volume is in milliliters, you need to convert it to liters. Therefore, 200.0 mL = 0.2000 L.

Step 3: Calculate the molarity. Molarity is the number of moles of solute divided by volume of the solution in liters. Hence, M = 0.0678 mol / 0.2000 L = 0.339 M. Therefore, the molarity of Pb(NO3)2(aq) solution is 0.339 M.

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Answer:

pH of Bfr = 4.83

Explanation:

The pH of the buffer solution prepared from acetic acid and calcium acetate in sufficient water is 4.84.

H(Ac) = 0.4/1.4L = 0.2857 M

Ca(OH)₂ = 0.25/1.4L = 0.17857 M

= 0.2857 M H(Ac)  + 0.17857 M Ca(OH)₂

= 0.2857 M H(Ac) + 2(0.17857) OH⁻

=  0.2857 M H(Ac) + 0.3571 M OH⁻

H(Ac) ⇄ H⁺   OH⁻

0.2857 M   ⇄ 0.3571 M

[tex]H^+ = \frac{[K_a][H(_{AC}]}{OH^-} \\\\H^+ = \frac{(1.8 \times 10^{-5}) (0.2857)}{(0.3571)} \\\\H^+ = 1.44\times 10^{-5}[/tex]

pH = -Log(H⁺)

pH = - Log(1.44 x 10⁻⁵)

pH = 4.84

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Answer:

(R) - hexyl acetate

Explanation:

Hello,

This reacción is a nucleophilic substitution SN2.

The configuration (s), means that the groups around the chiral carbon are organized appose to the clock hands movement. But when the reaction happens, these configurations become an (r) configuration, it means the groups around the chiral carbon organize according to the clock hands movement.

Generally, these reactions are related to nucleophilic species, an example is the ion acetate, a conjugated acid which is a weak nucleophilic, for this reason, the transition state is more energetic, it means, less stable than if the reaction occurs with a strong nucleophilic.

Look the image to compare the two configurations of the reactant and product.

An SN2 reaction between (S)-2-bromohexane and acetate ion leads to the formation of (R)-2-acetoxypentane due to the backside attack characteristic of such reactions, resulting in an inversion of the original stereochemistry around the chiral center.

The student's question involves a nucleophilic substitution (specifically an SN2 reaction) where the nucleophile, acetate ion (CH3CO2-), replaces the bromide ion in (S)-2-bromohexane. Under the assumption of inversion of configuration, the product will be (R)-2-acetoxypentane. The initial (S)-2-bromohexane has a specific three-dimensional arrangement with the bromine atom attached to the second carbon in a configuration that is opposite to that of the other substituents when viewed in a certain manner.

In an SN2 reaction, the nucleophile attacks from the opposite side of the leaving group, leading to an inversion of the stereochemistry at the carbon center undergoing the substitution. Therefore, the product will be the R-enantiomer of 2-acetoxypentane since the attacking acetate ion approaches from the side opposite to the leaving bromide ion, flipping the configuration.

The stereochemical outcome of an SN2 reaction is critical in organic synthesis as it controls the formation of specific enantiomers of chiral molecules.

Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .

Explanation: The pH of the buffer solution is calculated by using Handerson equation:

[tex]pH=pKa+log(\frac{base}{acid})[/tex]

Let's calculate the moles of acid and base(salt) present in the original buffer.

mL are converted to L and then multiplied by molarity to get the moles.

[tex]525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})[/tex]

= [tex]0.2625molHCHO_2[/tex]

[tex]475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})[/tex]

= [tex]0.2375molNaCHO_2[/tex]

Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L

Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.

pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.

[tex]pH=3.74+log(\frac{0.2375}{0.2625})[/tex]

pH = 3.74 - 0.04

pH = 3.70

Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.

[tex]85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})[/tex]

= [tex]0.0223molHCHO_2[/tex]

[tex]85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})[/tex]

= [tex]0.0202molNaCHO_2[/tex]

Now, let's calculate the moles of HCl added to the buffer.

[tex]8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})[/tex]

= 0.00129 mol HCl

This added HCl reacts with base(sodium formate) and formic acid is produced.

So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:

[tex]H^++CHO_2^-\rightarrow HCHO_2[/tex]

moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol

moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol

Let's plug in the values again in the Handerson equation and calculate the pH:

[tex]pH=3.74+log(\frac{0.01891}{0.02359})[/tex]

pH = 3.74 - 0.096

pH = 3.64

Answer: The given statement is false.

Explanation:

Precipitation reaction is defined as the chemical reaction in which two aqueous solution upon mixing together results in the formation of an insoluble solid.

For example, [tex]NaCl(aq) + AgNO_{3} \rightarrow NaNO_{3}(aq) + AgCl(s)[/tex]

Here AgCl is present in solid state so, it is the precipitate.

But it is not necessarily true that two aqueous solutions will always result in the formation of a precipitate.

For example, [tex]NaCl(aq) + KNO_{3}(aq) \rightarrow KCl(aq) + NaNO_{3}(aq)[/tex]

Hence, we can conclude that the statement precipitation reactions always occur when two aqueous solutions are mixed, is false.

The statement is false; precipitation reactions occur when two aqueous solutions form an insoluble product. Whether a reaction occurs depends on the solubility rules of the compounds formed.

The statement that precipitation reactions always occur when two aqueous solutions are mixed is false. A precipitation reaction occurs when two solutions are mixed and an insoluble product, the precipitate, forms. Whether a precipitation reaction occurs depends on the solubility rules of the ionic compounds formed during the mixing of the two solutions. If none of the possible combinations result in an insoluble product, then no precipitation will occur. This is described by solubility rules, which predict the solubility of different ionic compounds in water.

The percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.

We can use the following formula to determine the percentage volume change brought about by the allotropic conversion of iron (Fe) from bcc (α phase) to fcc ( γ phase):

Percentage volume change = [(Vγ - Vα) / Vα] * 100

Where

Vγ is the volume of the FCC phase and

Vα is the volume of the BCC phase.

We can use the following formula to determine the volume:

Volume = [tex](4/3) * \pi * r^3[/tex]

Where r is the radius of the atom.

We can calculate the volume as the atomic radius of Fe changes from RBCC = 0.12584 nm to RFCC = 0.12894 nm:

Vα = (4/3) * π *[tex](RBCC^3)[/tex]

Vγ = (4/3) * π * [tex](RFCC^3)[/tex]

Now that the numbers have been substituted, we can determine the percentage change in volume:

Percentage volume change = [(Vγ - Vα) / Vα] * 100

= [(Vγ / Vα) - 1] * 100

= [(Vγ / Vα) - 1] * 100

Vα = (4/3) * π * [tex](0.12584^3)[/tex] = 0.00167709 [tex]nm^3[/tex]

Vγ = (4/3) * π * [tex](0.12894^3)[/tex] = 0.00178565 [tex]nm^3[/tex]

Percentage volume change = [(0.00178565 / 0.00167709) - 1] * 100

= [(1.06494) - 1] * 100

= 0.06494 * 100

= 6.494%

Hence, the percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.

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The percentage volume change associated with given reaction is 7.6 %.

We can calculate the percentage volume change using the formula:

[tex]\% \text { Volume Change }=\frac{V_{\text {final }}-V_{\text {initial }}}{V_{\text {initial }}} \times 100[/tex]

The volume of a unit cell can be calculated using the formula V = [tex]a^3[/tex] , where a is the lattice parameter.

For the BCC (α phase), the initial volume ([tex]V_{\text {initial }}[/tex]) is given by [tex]a_{\mathrm{BCC}}^3[/tex], and for the FCC (γ phase), the final volume ([tex]V_{\text {final }}[/tex]) is given by [tex]a_{\mathrm{FCC}}^3[/tex].

Let's calculate it:

Given that a(BCC) is the lattice parameter for BCC phase (0.12584 nm) and a(FCC) is the lattice parameter for FCC phase (0.12894 nm).

[tex]\% \text { Volume Change }=\frac{(0.12894 \mathrm{~nm})^3-(0.12584 \mathrm{~nm})^3}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]

[tex]\% \text { Volume Change }=\frac{\left(0.0001509298202 \mathrm{~nm}^3\right)}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]

%Volume Change ≈ 0.076×100

%Volume Change ≈ 7.6 %

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}[/tex]

where,

[tex]p^o[/tex] = vapor pressure of pure solvent  (water) = 23.76 mmHg

[tex]p_s[/tex] = vapor pressure of solution= ?

[tex]w_2[/tex] = mass of solute  (sucrose) = 12.25 g

[tex]w_1[/tex] = mass of solvent  (water) = 176.3 g

[tex]M_1[/tex] = molar mass of solvent (water) = 18.02 g/mole

[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

[tex]\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}[/tex]

[tex]p_s=23.67mmHg[/tex]

Therefore, the vapor pressure of solution is 23.67 mmHg.

Final answer:

The vapor pressure of a solution made by dissolving 12.25 grams of sucrose in 176.3 grams of water at 25 °C is calculated as 23.68 mm Hg, using Raoult's Law and the vapor pressure of pure water (23.76 mm Hg) as the basis.

Explanation:

Calculation of Vapor Pressure of a Sucrose Solution

To calculate the vapor pressure of a solution made by dissolving 12.25 grams of sucrose in 176.3 grams of water at 25 °C, where the vapor pressure of pure water is 23.76 mm Hg, we use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution (Πsolvent) is equal to the vapor pressure of the pure solvent (Πpure solvent) times the mole fraction of the solvent (χsolvent) in the solution.

First, calculate the moles of sucrose (C12H22O11): Moles = 12.25 g / 342.3 g/mol.

Then, calculate the moles of water (H2O): Moles = 176.3 g / 18.02 g/mol.

Next, compute the mole fraction of water: χH2O = moles of H2O / (moles of H2O + moles of C12H22O11).

Finally, calculate vapor pressure of the solution: Πsolution = Πpure water * χH2O.

Working through the calculations:

Moles of sucrose = 12.25 / 342.3 = 0.0358 mol.
Moles of water = 176.3 / 18.02 = 9.785 mol.
Mole fraction of water = 9.785 / (9.785 + 0.0358) = 0.99636.
Vapor pressure of the solution = 23.76 mm Hg * 0.99636 = 23.68 mm Hg.

Thus, the vapor pressure of the sucrose solution at 25 °C is 23.68 mm Hg.

Answer: The empirical formula of compound is [tex]C_4H_{12}Si[/tex].

Explanation:

Mass of Sample= 0.702 g

Mass of [tex]CO_2[/tex] = 1.4 g

Mass of [tex]H_2O[/tex] = 0.86 g

Mass of [tex]SiO_2[/tex] = 0.478 g  

First we have to calculate moles of[tex]CO_2[/tex], [tex]H_2O[/tex] and [tex]SiO_2[/tex] formed.

1. Moles of [tex]CO_2=\frac{1.4g}{44g/mol}=0.032mol[/tex]

Now , Moles of carbon == Moles of [tex]CO_2[/tex] = 0.032

2.  Moles of [tex]H_2O=\frac{0.86g}{18g/mol}[/tex]=0.048mol​​​

Now , Moles of hydrogen = [tex]2\times[/tex] Moles of [tex]H_2O[/tex] =[tex]2\times 0.048=0.096mol[/tex]

3.  Moles of [tex]SiO_2=\frac{0.478g}{60g/mol}=0.008[/tex] mol

Now , Moles of silicon = Moles of [tex]SiO_2[/tex] = 0.008 moles

Therefore, the ratio of number of moles of C : H : Si is  = 0.032 : 0.096 : 0.008

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C= [tex]\frac{0.032}{0.008}=4[/tex]

For H =[tex]\frac{0.096}{0.008}=12[/tex]

For Si=[tex]\frac{0.008}{0.008}=1[/tex]

Thus, C: H: Si = 4 : 12 : 1

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is [tex]C_4H_{12}Si[/tex].

Final answer:

To find the empirical formula of the compound, calculate the moles of each element from the given masses of CO2, H2O, and SiO2. The empirical formula for this compound is CH4Si.

Explanation:

To determine the empirical formula of the compound, first, calculate the moles of each element using the given masses of CO2, H2O, and SiO2. Next, find the ratio of the moles of each element to each other and simplify if necessary to get the empirical formula. In this case, the empirical formula of the compound is **CH4Si**.

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

[tex]U_{2}-U_{1}=Q-W[/tex]

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

[tex]dw=Pdv[/tex]

The pressure is constant, so:  [tex]w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}[/tex] (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

[tex]U_{2}-U_{1}=500-(3*75)=275kJ[/tex]

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

[tex]P_{1}V_{1}=nRT_{1}[/tex]

[tex]P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}[/tex]  

Subtracting the first from the second:

[tex]1.5P_{1}V_{1}=nR(T_{2}-T_{1})[/tex]

Isolating [tex]T_{2}[/tex]:

[tex]T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}[/tex]

Assuming that it is water steam, n=0.1666 kmol

[tex]V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}[/tex]

[tex]T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76[/tex] ºC

Answer: The concentration of NaCl in 1.00 L of solution is 0.0076 M.

Explanation:

To calculate the molarity of the concentrated solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]  

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution

We are given:

Conversion factor used: 1L = 1000 mL

[tex]M_1=?M\\V_1=1L=1000mL\\M_2=0.076M\\V_2=100mL[/tex]

Putting values in above equation, we get:

[tex]M_1\times 1000=0.076\times 100\\\\M_1=0.0076M[/tex]

Hence, the concentration of NaCl in 1.00 L of solution is 0.0076 M.

To find the amount of water produced when 2.5 mol of O2 is consumed in the combustion of propane, we can use stoichiometry. The combustion of 2.5 mol of O2 will produce 2 mol of H2O.

The balanced equation for the combustion of propane is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

To find the amount of water produced when 2.5 mol of O2 is consumed, we need to use stoichiometry. Since the ratio of O2 to H2O is 5:4, we can set up the following proportion:

2.5 mol O2 / 5 mol O2 = x mol H2O / 4 mol H2O

Solving for x, we find that x = 2 mol H2O. Therefore, the combustion of 2.5 mol of O2 will produce 2 mol of H2O.

Answer:

5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7

Explanation:

Hello,

In this case, the decreasing basicity order depends on the metal contained into the oxide, in such a way, on the attached picture you will find how the basicity increases leftwards and downwards on the periodic table, as the metallic character does.

In this manner, as the cesium in the Cs₂O is the leftest atom, it is the most basic, next MgO, afterwards SnO₂, then P₄O₁₀ and finally Cl₂O₇, therefore, the order is: 5) Cs₂O> MgO > SnO₂> P₄O₁₀> Cl₂O₇.

Best regards.

The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.

Metals are known to form basic oxides. The oxides of metals react with acids to form salt and water. This is one of the properties of metals.

However, the basicity of metallic oxides decreases across the period. As you go from left to right the metal oxides become less basic. Recall that the oxides of metals are acidic. The react with bases to form salts and water.

The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.

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Answer:

The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].

Explanation:

Mass of copper ball = 50.0 g

Moles of copper = [tex]\frac{50.0 g}{63.5 g/mol}[/tex]

1 mole = [tex]N_A=6.022\times 10^{23}[/tex]

Number of copper atoms =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} [/tex]

1 atom of copper has 29 protons

Total number of protons in 50.0 g of copper =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} \times 29=1.3751\times 10^{25}[/tex]

Since an atom is a neutral specie which means number of protons are equal to number of electrons.

Total number of electrons = [tex]1.3751\times 10^{25}[/tex]....(1)

Net charge on the copper ball = [tex]2.00/mu C=2.00\times 10^{-6} C[/tex]

Q=Ne

Q = Total charge

N = Number of electrons

e = charge on an electron = [tex]1.602\times 10^{-19} C[/tex]

[tex]2.00\times 10^{-6} C=N\times 1.602\times 10^{-19} C[/tex]

[tex]N =1.248\times 10^{13} [/tex]

Total number of electrons removed = N = [tex]1.248\times 10^{13} [/tex]

Fraction of the copper’s electrons has been removed:

[tex]\frac{\text{Number of electrons removed}}{\text{Total electrons}}[/tex]

[tex]\frac{1.248\times 10^{13}}{1.3751\times 10^{25}}=9.076\times 10^{-13}[/tex]

The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].

To calculate the fraction of electrons removed from a 50.0 g ball of copper with a 2.00 µC charge, we determine the total number of electrons in the ball and then find the number corresponding to the charge. We conclude that approximately 9.083 x 10−11% of the copper's electrons have been removed.

To determine the fraction of copper's electrons that have been removed, we'll need to calculate the total number of electrons in the 50.0 g ball of copper and then see how many electrons correspond to a charge of 2.00 µC.

First, we calculate the number of moles of copper in the 50.0 g ball. With an atomic mass of 63.5 g/mol for copper, we have:

Number of moles = 50.0 g / 63.5 g/mol = 0.7874 moles

Next, using Avogadro's number (6.022 x 1023 atoms/mol), we find the number of copper atoms:

Number of copper atoms = 0.7874 moles x (6.022 x 1023 atoms/mol) = 4.739 x 1022 atoms

Since each copper atom contributes 29 electrons, the total number of electrons in the ball is:

Total number of electrons = 29 x 4.739 x 1022 atoms = 1.374 x 1024 electrons

To find the number of electrons that corresponds to a charge of 2.00 µC, we use the electron charge (1.602 x 10−19 C/electron):

Number of electrons removed = 2.00 µC / (1.602 x 10−19 C/electron) = 2.00 x 10−6 C / (1.602 x 10−19 C/electron) = 1.248 x 1013 electrons

Now, we find the fraction of electrons removed:

Fraction = Number of electrons removed / Total number of electrons = 1.248 x 1013 / 1.374 x 1024 ≈ 9.083 x 10−11%

So, approximately 9.083 x 10−11% of the copper's electrons have been removed.

Answer:

the compound is hexadeca-6,10-diyne.

Explanation:

As given that the compound has two triple bonds and it is giving carboxylic acids on ozonolyis so this is an oxidative ozonolysis.

We are getting only two kinds of products, it means the triple bonds are located at symmetrical position.

The middle carbon chain must have four carbons and the side chains must have six carbons each,

The structure of hydrocarbon is shown in figure.

Answer:

The answer would be Lead.

Explanation:

The problem is solved using system of equations. Considering the given fat and cholesterol content of each soup, we create two equations and solve for x (Chunky Beef soup servings) and y (Chunky Sirloin Burger soup servings), finding that 4 servings of each soup are needed.

This problem can be solved using system of equations. Let's define x as the number of servings of Chunky Beef soup and y as the number of servings of the Chunky Sirloin Burger soup. Considering the fat content from both types of soup, we get the equation 2.5x + 7y = 29. The cholesterol content in both soups is the same (15mg per serving), yielding a second equation 15x + 15y = 120. The second equation simplifies to x + y = 8, meaning that we will have a total of 8 servings.

By solving the system of equations, we find that x (the number of cups of Chunky Beef Soup) is 4 and y (the number of cups of the Chunky Sirloin Burger soup) is also 4.

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The solution involves solving a system of equations. The calculation shows that to get 8 servings of soup with 29 g of fat and 120mg of cholesterol, we should mix approximately 3.33 cups of the Chunky Beef soup and 4.67 cups of the Chunky Sirloin Burger soup.

This problem can be approached using a system of linear equations. Assume x is the number of cups of Campbell's Chunky Beef soup and y is the number of Campbell's Chunky Sirloin Burger soup.

From the question, we have:
1. x + y = 8 (since there are 8 servings total)
2. 2.5x + 7y = 29 (calculating total fat in grams)
3. 15x + 15y = 120 (calculating total cholesterol in milligrams).

The third equation simplifies to x + y = 8. This is equivalent to the first equation, so it doesn't provide any new information. We can therefore solve the first and second equations simultaneously.

Subtract the first equation from the second to get 4.5y = 21, from which y = 21/4.5 = 4.67 (approximately). Substituting y = 4.67 into the first equation, we get x = 8 - 4.67 = 3.33. Therefore, to get 8 servings of soup with 29 g of fat and 120 mg of cholesterol, we should mix approximately 3.33 cups of Campbell's Chunky Beef soup and 4.67 cups of Campbell's Chunky Sirloin Burger soup.

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Explanation:

[tex]HBr(aq) + LiOH(aq) \rightarrow LiBr(aq)+H_2O(l)[/tex]

a)Molarity of LiOH,[tex]M_1[/tex] = 0.0024 M

Volume of LiOH = [tex]V_1[/tex]

Molarity of HBr,[tex]M_2[/tex] = 0.0026 M

Volume of HBr = [tex]V_2=22mL=[/tex]

According to reaction , 1 mol of LiOH neutralizes 1 mol of HBr.

[tex]M_1V_1=M_2V_2[/tex]

[tex]V_1=\frac{0.0026 M\times 22mL}{0.0024 M}=23.8333 mL[/tex]

[tex]M_1=\frac{\text{Molesof LiOH}}{V_1}[/tex]

Moles of LiOH = [tex]0.0024 mol/L\times 23.8333 ml=0.0571 mol[/tex]

Mass of 0.0571 mol of LiOH:

[tex]0.0571 mol\times 2 g/mol=1.3704 g[/tex]

1.3704 g of 0.0024 M LiOH that would neutralize 22 mL of 0.0026 M HBr.

b) According to reaction , 1 mol LiOH gives 1 mol of LiBr.

Then ,0.0571 mol of LiOH will give:

[tex]\frac{1}{1}\times 0.0571 mol=0.0571 mol[/tex] of LiBr

0.0571 moles of salt are produced in the reaction

c) Moles of salt = 0.0571 mol

Volume of the solution = 22 ml+ 23.8333 mL= 45.8333 mL = 0.0458333 L

Molar concentration of the salt:

[tex]{LiBr}=\frac{0.0571 mol}{0.0458333 L}=1.2458 mol/L[/tex]

1.2458 mol/L is the molar concentration of the salt after the reaction is complete.

Hydrobromic acid and lithium hydroxide react to form lithium bromide and water. 23.8 mL of 0.0024 M LiOH would neutralize 22 mL of 0.0026 M HBr, producing 0.0000572 mol of LiBr. The molar concentration of the salt solution after the reaction = 0.00125 M.

The balanced chemical equation for the reaction between hydrobromic acid (HBr) and lithium hydroxide (LiOH) is given by: HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l).

Considering the balanced equation, we observe a ratio of 1:1 between HBr and LiOH. In (a), to calculate the volume of 0.0024 M LiOH necessary to neutralize 22 mL of 0.0026 M HBr, we use the concept of molarity (M) which is equal to mol/L. Therefore, the solution of 0.0024 M LiOH has 0.0024 mol of LiOH per L. As the ratio is 1:1, we also need 0.0026 mol/L * 0.022L = 0.0000572 mol of LiOH, which means we need 0.0000572 mol / 0.0024 mol/L = 0.0238L or 23.8 mL of LiOH.

In (b), as the reaction produces 1 mol of LiBr(salt) for every mole of HBr or LiOH, there will be the same amount of salt produced, so 0.0000572 mol of salt is produced.

In (c), the molar concentration of salt solution will be the number of moles of the salt divided by total volume of the solution. Assuming volumes of reactants add up, total volume = 0.022 L + 0.0238 L = 0.0458 L. Therefore, the molar concentration = 0.0000572mol / 0.0458L = 0.00125 M.

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Answer:

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Explanation:

Animals take in oxygen and breathe out carbon dioxide during cellular respiration. The reaction for the metabolism of the food in the animal body is:

[tex]C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O + Energy[/tex]

As can be seen from the reaction stoichiometry that:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

Also,

Given :

Mass of carbon dioxide gas = 120 g

Molar mass of carbon dioxide gas = 44 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus, moles of carbon dioxide are:

[tex]moles_{CO_2} = \frac{120 g}{44 g/mol}[/tex]

[tex]moles_{CO_2} = 2.7273 mol[/tex]

As mentioned:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

1 mole of carbon dioxide gas can be produced from 1/6 mole of oxygen gas in the process of metabolism of glucose.

2.7273 mole of carbon dioxide gas can be produced from [tex] \frac{1}{6} \times 2.7273[/tex] moles of oxygen gas in the process of metabolism of glucose.

Thus, moles of oxygen gas needed = 0.4545 moles

Molar mass of oxygen gas = 32 g/mol

The mass of oxygen gas can be find out by using mole formula as:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Mass\ of\ oxygen\ gas = Moles \times Molar mass}[/tex]

[tex]Mass\ of\ oxygen\ gas = 0.4545 \times 32}[/tex]

[tex]Mass\ of\ oxygen\ gas = 14.544 g[/tex]

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Answer:

87.3 g

Explanation:

The cellular respiration can be represented through the following equation.

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

We can establish the following relations:

The mass of O₂ that produces 120 g of CO₂ is:

[tex]120gCO_{2}.\frac{1molCO_{2}}{44.01gCO_{2}} .\frac{6molO_{2}}{6molCO_{2}} .\frac{32.00gO_{2}}{1molO_{2}} =87.3gO_{2}[/tex]

Answer:

The answer is B

[The extracting solvent should be non-toxic and readily available]

Explanation:

The extracting solvent should not pose a risk to life (or the risk have to be minimum) and be available in large quantity.

A. Most of the extracting solvents are volatile, and is a good think because you can remove them by distillation.

C. The extraction solvent should not react with the solute, because you lose the extracted compound.

D. The extraction solvent should not be miscible with water. Usually you extract compounds organic compounds from water using an appropriate extracting solvent.

The correct attribute for an extracting solvent is that it should be non-toxic and readily available, making (b) the right answer. Volatility, reactivity with the solute, and miscibility with water are undesirable features for an extraction solvent.

An extracting solvent is used in liquid-liquid extraction to separate compounds based on their solubility properties. The ideal characteristics of an extracting solvent include being non-toxic, readily available, not miscible with water, and having a different density than water to facilitate easy separation. For example, diethyl ether is an effective extracting solvent because of its ability to dissolve non-ionic organic compounds, immiscibility with water, and the fact that ionic compounds are generally insoluble in it. In contrast, Toluene (C₆H₅-CH₃) is a nonpolar solvent which is ideal for dissolving nonpolar substances like octane , but not polar substances such as water (H₂O) or ionic compounds like sodium sulfate .

In selecting the correct answer to the student's question, the attribute that an extracting solvent should have is that it should be non-toxic and readily available. Solvents that are volatile, reactive with the solute, or miscible with water are not ideal for extraction purposes. Therefore, the correct choice is (b).

The concentration of NO3- ions in solution C, which is made from a mixture of NaCl and AgNO3 solutions, is calculated to be 0.80 M.

To calculate the concentration of the NO3- in solution C, we consider both AgNO3 and NaCl. In AgNO3( aq), each molecule yields an Ag+ and a NO3-. The total amount of moles of NO3- from this solution can be calculated by multiplying the volume of the solution(2.00 L) by its concentration(2.00 M). This gives 4.00 moles of NO3-.

As for NaCl(aq), it yields Na+ and Cl- ions when dissolved but no NO3- ions. So, NO3- ions will only come from solution B in this case.

The total volume of solution C is the volume of solution A plus the volume of solution B, that is, 3.00L + 2.00L = 5.00L. We can now calculate the concentration of NO3- ions in solution C by dividing the total moles of NO3- (4.00 moles) by the total volume of solution C (5.00 L). Therefore, the concentration of the NO3- ion in solution C is 4.00 moles / 5.00 L = 0.80 M.

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The concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.

The concentration of [tex]NO3^-[/tex]ions in solution C can be calculated by considering the moles of [tex]NO3^-[/tex] present in solution B and the final volume of solution C after mixing.

First, let's calculate the moles of [tex]NO3^-[/tex] in solution B:

Moles of [tex]NO3^-[/tex] = Concentration of [tex]NO3^-[/tex] in solution B × Volume of solution B

Moles of  [tex]NO3^-[/tex]= 2.00 M × 2.00 L = 4.00 moles

Now, let's find the final volume of solution C after mixing solutions A and B:

Volume of solution C = Volume of solution A + Volume of solution B

Volume of solution C = 3.00 L + 2.00 L = 5.00 L

Next, we calculate the concentration of [tex]NO3^-[/tex] in solution C:

Concentration of [tex]NO3^-[/tex] in solution C = Moles of [tex]NO3^-[/tex]/ Volume of solution C

Concentration of [tex]NO3^-[/tex] in solution C = 4.00 moles / 5.00 L = 0.800 M

Therefore, the concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.

The answer is: 0.800.

Answer: n-Butanol are converted using SN2 and tert-butanol is converted using SN1

Explanation: For the conversion of n-butanol into butyl chloride using Hydrogen Chloride the reaction would follow SN2 mechanism.

SN2 reaction mechanism occurs only in the case of primary substrates as it is a one step mechanism that happens in a concerted manner. It involves backside attack of nucleophile on the substrate such that the nucleophile attacks from the back side and leaving group leaves from the front side.

In this reaction since hydroxy group (OH) is not a good leaving group hence firstly we need to convert it into a good leaving group. When we treat n-butanol with HCl hydroxy group is protonated and now it turns into a good leaving group as it can leave as H₂O.

Cl⁻  here acts as nucleophile and now attacks the primary carbon center from the back side which contains the protonated hydroxy group as a leaving group.

In the case of tertiary butanol the reaction follows SN1  mechanism and it is 2 step mechanism.

In the first step hydroxy group is protonated and as it becomes a good leaving group it leaves and leads to the formation of a stable tertiary carbocation as an intermediate.

In the second step this intermediate carbocation is attacked by the Cl⁻  nucleophile which leads to the formation of tertiary butyl chloride.

Kindly find in attachment the reaction mechanism for both the reactions.

Answer: The volume of the container is [tex]2.8497m^3[/tex]

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

[tex]PV=nRT[/tex]

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g   (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]

T = temperature of container = [tex]200^oC=[200+273]K=473K[/tex]

Putting values in above equation, we get:

[tex]200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L[/tex]

Converting this into cubic meter, we use the conversion factor:

[tex]1m^3=1000L[/tex]

So, [tex]\Rightarrow \frac{1m^3}{1000L}\times 2849.7L[/tex]

[tex]\Rightarrow 2.8497m^3[/tex]

Hence, the volume of the container is [tex]2.8497m^3[/tex]

Answer:

Concentration of sodium ions in final solution is 0.1428 mol/L.

Concentration of nitrate ions in final solution is 0.2857 mol/L.

Explanation:

Concentration =[tex]\frac{\text{Moles of compound}}{\text{Volume of the solution (L)}}[/tex]

[tex]Na_2CrO_4(aq)+2AgNO_3(aq)\rightarrow Ag_2CrO_4(s)+2NaNO_3(aq)[/tex]

Moles of silver nitrate in 10 mL of 1 M silver nitrate.

Volume of the  silver nitrate solution = 10 mL = 0.010 L

[tex]1 M =\frac{\text{Moles of}Ag_2CrO_4}{\text{Volume of the solution in L}}[/tex]

Moles of silver nitrate =[tex]1 M\times 0.010 L=0.010 mol[/tex]

Moles of sodium chromate in 25 mL of 0.1 M silver nitrate.

Volume of the  silver chromate solution = 25 mL = 0.025 L

[tex]0.1 M =\frac{\text{Moles of}Na_2CrO_4}{\text{Volume of the solution in L}}[/tex]

Moles of sodium nitrate =[tex]0.1 M\times 0.025 L=0.0025 mol[/tex]

According to reaction, 1 mole of sodium chromate reacts with 2 mole of silver nitrate.

Then , 0.0025 mol of sodium chromate will react with :

[tex]\frac{1}{2}\times 0.0025 mol=0.00125 mol[/tex] of silver nitrate.

1 mol of sodium chromate gives 2 mol of sodium ions and 1 mol of chromate ions.

Then 0.0025 mol of sodium chromate will give 0.0050 mol of sodium ions.

Volume of the solution after mixing =

10 mL + 25 mL = 35 mL =0.035 L

Concentration of sodium ions in final solution:

[tex][Na^+]=\frac{0.0050 mol}{0.035 mL}=0.1428 mol/L[/tex]

Concentration of sodium ions = 0.1428 mol/L

1 mole of silver nitrate gives 1 mol of silver ion and 1 mole nitrate ion

Then 0.010 moles of silver nitrate will give 0.010 moles of nitrate ions.

Concentration of nitrate ion in the final solution:

[tex][NO_3^{-}]=\frac{0.010 mol}{0.035 L}=0.2857 mol/L[/tex]

Concentration of nitrate ions = 0.2857 mol/L

The concentration of sodium ion after the two solutions are combined is 0.1428M.

Concentration will be calculated in terms of molarity as:

M = n/V, where

n = no. of moles

V = volume

Given chemical reaction is:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

Given volume of AgNO₃ = 10mL = 0.010L

Molarity of AgNO₃ = 1M

Moles of AgNO₃ = 0.010×1 = 0.010 moles

Given volume of Na₂CrO₄ = 25mL = 0.025L

Molarity of Na₂CrO₄ = 0.1M

Moles of Na₂CrO₄ = 0.025×0.1 = 0.0025 moles

From the stoichiometry of the reaction it is clear that:

1 mole of Na₂CrO₄ = produce 2 moles of sodium ions

0.0025 moles of Na₂CrO₄ = produce 2×0.0025=0.0050 moles of sodium ions

Total volume of final solution = 25mL + 10mL = 35mL = 0.035L

Now we calculate the concentration in terms of molarity of sodium ions as:
M = 0.0050/0.035 = 0.1428M

Hence, the concentration of sodium ions is 0.1428M.

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The balanced chemical reactions are:

[tex]AgNO{_3}(aq) \ + \ NaCl(aq) \rightarrow \ AgCl(s) \ + NaNO_{3}(aq)\\2K_{3}PO_{4}(aq) \ + 3MgCl_{2}(aq) \rightarrow \ 6KCl(aq) \ + Mg_{3}(PO_{4})_{2}(s)[/tex]

Further Explanation:

The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.

For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:

Therefore, the products will have the following characteristics:

For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:

Hence, the products of the second reaction will have the following characteristics:

Insoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as (aq).

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Keywords: solubility rules, precipitation reaction

Here are the balanced precipitation reactions with physical states: 1. AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq), 2. K3PO4(aq) + MgCl2(aq) -> 3KCl(aq) + Mg3(PO4)2(s)

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