Answer:
The correct answer is that letter. (μ)
Explanation:
Hello!
According to the international system of units, we use prefixes to indicate the number of places we ran the point.
The letter μ indicates [tex]10^{-6}[/tex]
The correct answer is that letter. (μ)
It is not in the options but it is correct.
Calculate the volume of 38.0 g of carbon dioxide at STP. Enter your answer in the box provided. L
Answer:
19.3 L
Explanation:
V= n × 22.4
where V is volume and n is moles
First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01
n= m/ MM
n= 38/ 44.01
n= 0.86344012724
V= 0.86344012724 × 22.4
V= 19.3410588502 L
V= 19.3 L
Components A and B form ideal solution. At 350 K, a liquid mixture containing 40% (mole) A is in equilibrium with a vapour containing 70% (mole) A. If the vapour pressure of A at 350 K is 70 kPa, what is the vapour pressure of B? (b) 20 kPa (d) 12 kPa (а) 25 kPa (c) 40 kPa
Answer : The correct option is, (b) 20 kPa
Explanation :
The Raoult's law for liquid phase is:
[tex]p_A=x_A\times p^o_A[/tex] .............(1)
where,
[tex]p_A[/tex] = partial vapor pressure of A
[tex]p^o_A[/tex] = vapor pressure of pure substance A
[tex]x_A[/tex] = mole fraction of A
The Raoult's law for vapor phase is:
[tex]p_A=y_A\times p_T[/tex] .............(2)
where,
[tex]p_A[/tex] = partial vapor pressure of A
[tex]p_T[/tex] = total pressure of the mixture
[tex]y_A[/tex] = mole fraction of A
Now comparing equation 1 and 2, we get:
[tex]x_A\times p^o_A=y_A\times p_T[/tex]
[tex]p_T=\frac{x_A\times p^o_A}{y_A}[/tex] ............(3)
First we have to calculate the total pressure of the mixture.
Given:
[tex]x_A=0.4[/tex] and [tex]x_B=1-x_A=1-0.4=0.6[/tex]
[tex]y_A=0.7[/tex] and [tex]y_B=1-y_A=1-0.7=0.3[/tex]
[tex]p^o_A=70kPa[/tex]
Now put all the given values in equation 3, we get:
[tex]p_T=\frac{0.4\times 70kPa}{0.7}=40kPa[/tex]
Now we have to calculate the vapor pressure of B.
Formula used :
[tex]x_B\times p^o_B=y_B\times p_T[/tex]
[tex]p^o_B=\frac{y_B\times p_T}{x_B}[/tex]
Now put all the given values in this formula, we get:
[tex]p^o_B=\frac{0.3\times 40kPa}{0.6}=20kPa[/tex]
Therefore, the vapor pressure of B is 20 kPa.
When measuring a solid metal block at constant temperature, which measurement will change in numerical value depending on the location where it is taken? A) weight B) length C) volume D) mass
Answer:
A) weight
Explanation:
The weight is the force that gravity makes in a body, and it is given by the equation:
W = mxg
Where W is the weight, m is the mass, and g is the gravity in the local. So, the length, the volume, and the mass are constant and don't depend on the local that is measured, but the weight depends on the local gravity.
How do you calculate the pH of a 1.6 M KOH solution?
Answer:
14.20
Explanation:
The pH can be calculated as:
KOH is a strong base and it will dissociate completely in the solution as:
[tex]KOH\rightarrow K^++OH^-[/tex]
According to the equation and also mentioned that KOH will dissociate completely. So,
Molarity of the hydroxide ions furnished = Molarity of the KOH = 1.6 M
So,
pOH = -log[OH⁻] = -log(1.6) = -0.20
Also,
pH + pOH = 14
So, pH = 14 - ( - 0.20 ) = 14.20
What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
5.5 x 10-5
4.0 x 10-6
6.9 x 10-4
3.5 x 10-10
4.0 x 10-11
Answer:
Ka = [tex]4.04 \times 10^{-11}[/tex]
Explanation:
Initial concentration of weak acid = [tex]4.5 \times 10^{-4}\ M[/tex]
pH = 6.87
[tex]pH = -log[H^+][/tex]
[tex][H^+]=10^{-pH}[/tex]
[tex][H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M[/tex]
HA dissociated as:
[tex]HA \leftrightharpoons H^+ + A^{-}[/tex]
(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = [tex]1.35 \times 10^{-7}\ M[/tex]
[tex]Ka = \frac{[H^+][A^{-}]}{[HA]}[/tex]
[tex]Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}[/tex]
0.000000135 <<< 0.00045
[tex]Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}[/tex]
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
reactor: What is the maximum number of kg of NH3 that can be
produced?
Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of [tex]N_2[/tex] = 100 kg = 100000 g
Mass of [tex]H_2[/tex] = 100 kg = 100000 g
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles[/tex]
[tex]\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react with [tex]3571.43\times 3=10714.29[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react to give [tex]3571.43\times 2=7142.86[/tex] moles of [tex]NH_3[/tex]
Now we have to calculate the mass of [tex]NH_3[/tex]
[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]
[tex]\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg[/tex]
Therefore, the mass of ammonia produced can be, 121.429 kg
A balloon filled with helium gas at 1.00 atm occupies 12.9 L. What volume would the balloon occupy in the upper atmosphere, at a pressure of 0.19 atm and a constant temperature? Enter your answer in the box provided.
Answer:
67,9 L
Explanation:
Boyle's Law indicates that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of a gas, for a constant amount of gas we can write:
P1V1=P2V2
For the problem:
P1= 1 atm, V1= 12,9 L
P2=0,19 atm, V2=?
Therefore:
V2=P1V1/P2.................... V2=1 atm*12,9L/0,19 atm = 67,9 L
The balloon would occupy a volume of 67,9 L in the upper atmosphere.
How many significant figures does the rounded ans 101 cm - 99.1 cm = 1.9 cm (unrounded) After rounding, the answer should be reported to the and thus has 0.5 atm + 9.8 atm = 10.3 atm (unrounded) After rounding, the answer should be reported to the and thus has
Explanation:
Rules for counting significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.1) 101 cm - 99.1 cm = 1.9 cm ≈ 2.0 cm
1.9 cm ≈ 2.0 cm
There are two significant figures that is 2 and 0.
2) 0.5 atm + 9.8 atm = 10.3 atm ≈ 10.0 atm
There are three significant figures that is 3 , 0 and 0.
A gas of potassium chlorate molecules KClO3 all decompose into potassium chloride, KCl, and diatomic oxygen, O2. The products and reactants are in a closed container and can all be treated as ideal gases.A. Fill in the smallest possible integers that allows the stoichiometry of the reaction equation to be correct:__ KClO3 → ___ KCl ___ O2B. If there are N molecules of potassium chlorate in the initial state, how many product molecules are there
Answer: A. [tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. [tex]2.5\times N[/tex] molecules of product.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
A. The balanced reaction for decomposition of potassium chlorate is:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
According to stoichiometry :
2 moles of reactant give 5 moles of products
Thus [tex]2\times 6.023\times 10^{23}[/tex] molecules of reactant give [tex]5\times 6.023\times 10^{23}[/tex] molecules of product
Thus N molecules of reactant give= [tex]\frac{5\times 6.023\times 10^{23}}{2\times 6.023\times 10^{23}}\times N=\frac{5}{2}\times N[/tex] molecules of product.
Thus [tex]\frac{5}{2}\times N=2.5\times N[/tex] molecules of product are there.
Final answer:
The balanced chemical equation for the decomposition of potassium chlorate is 2 KClO3 → 2 KCl + 3 O2. If there are N molecules of potassium chlorate in the initial state, the number of product molecules of O2 can be calculated using the ratio of 2:3 in the equation.
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate is:
2 KClO3 → 2 KCl + 3 O2
The stoichiometry of the reaction equation can be balanced by filling in the smallest possible integers as follows:
A. 2 KClO3 → 2 KCl + 3 O2
B. If there are N molecules of potassium chlorate in the initial state, the number of product molecules can be calculated as follows:
The ratio of KClO3 to O2 in the balanced equation is 2:3. So, for every 2 molecules of KClO3 that decompose, 3 molecules of O2 are formed. Therefore, the number of O2 molecules produced is given by the equation:
(3 / 2) N = (3 / 2) × N molecules of O2
In 2006, a Russian team discovered an interesting molecule they called "sulflower" because of its shape and because it was based on sulfur. It is composed of 57.17% S and 42.83% C and has a molar mass of 448.70 g/mol. Determine the empirical and molecular formulas of "sulflower."
Answer: The empirical and molecular formula for the given organic compound is [tex]C_2S[/tex] and [tex]C_{16}S_8[/tex] respectively.
Explanation:
We are given:
Percentage of C = 42.83 %
Percentage of S = 57.17 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 42.83 g
Mass of S = 57.17 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.83g}{12g/mole}=3.57moles[/tex]
Moles of Sulfur = [tex]\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{57.17g}{32g/mole}=1.79moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.79 moles.
For Carbon = [tex]\frac{3.57}{1.79}=1.99\approx 2[/tex]
For Sulfur = [tex]\frac{1.79}{1.79}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : S = 2 : 1
Hence, the empirical formula for the given compound is [tex]C_2S_1=C_2S[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 448.70 g/mol
Mass of empirical formula = 56 g/mol
Putting values in above equation, we get:
[tex]n=\frac{448.70g/mol}{56g/mol}=8[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(8\times 2)}S_{(8\times 1)}=C_{16}S_8[/tex]
Thus, the empirical and molecular formula for the given compound is [tex]C_2S[/tex] and [tex]C_{16}S_8[/tex]
Final answer:
The empirical formula of 'sulflower' is CS2, found by using the given percentages to calculate the simplest whole number ratio of moles of carbon to sulfur. Dividing the compound's molar mass by the empirical formula's molar mass indicates that the molecular formula of 'sulflower' is C6S12.
Explanation:
To determine the empirical formula of the molecule known as 'sulflower', we can start by using the given percentages to find the moles of each element in the compound. Assuming we have 100 grams of the substance, we would have 57.17 grams of sulfur (S) and 42.83 grams of carbon (C). Using their molar masses (atomic sulfur is 32 g/mol and carbon is 12 g/mol), we can calculate the moles of each:
Sulfur: 57.17 g S × (1 mol S / 32 g) = 1.786 moles SCarbon: 42.83 g C × (1 mol C / 12 g) = 3.569 moles CTo find the empirical formula, we then find the simplest whole number ratio by dividing by the smallest number of moles, which gives us:
1.786 moles S / 1.786 moles = 1 S3.569 moles C / 1.786 moles = 2 CSo, the empirical formula is CS2. To determine the molecular formula, we use the molar mass of the empirical formula and divide the given molar mass of the compound by this value. The empirical formula mass of CS2 is (12.01 g/mol × 2) + (32.07 g/mol × 1) = 76.09 g/mol. The molecular formula is then found by dividing the compound's molar mass by the empirical formula's molar mass:
448.70 g/mol / 76.09 g/mol = 5.9 ≈ 6
Therefore, the molecular formula is C6S12, as the empirical formula must be multiplied by 6 to get the molecular formula.
A torsion balance has a sensitivity requirement (SR) of 4.5 mg. What is the MWQ of this balance if the maximum error permitted in using it is 3.6%?
Explanation:
MWQ means the minimum weighable quantity.
Mathematically, MWQ = [tex]\frac{Sensitivity}{1 - \text{fraction of accuracy}}[/tex]
or, MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]
It is given that sensitivity is 4.5 mg and maximum permitted error is 3.6%.
Therefore, fraction error = [tex]\frac{3.6}{100}[/tex] = 0.036
Hence, we will calculate MWQ as follows.
MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]
= [tex]\frac{4.5 mg}{0.036}[/tex]
= 125 mg
Thus, we can conclude that the MWQ of the given balance is 125 mg.
Maressa has 150 g of a NaCl solution that has concentration of 15.0 % NaCl (mm). How much NaCl is dissolved in that solution? O a. 0.05 O b. 1.08 Oc2.25 O d. 225 Oe. None of the above.
Answer:
I don't know if option e. is 22,5 g. because that is the correct answer for this. 15% NaCl (mm) gives you the information that in 100 g of solution, you have 15 g of solute. So, if in 100 g of solution you have 15 g of NaCl, in 150 g of solution, how much. Try the rule of three.
Explanation:
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−3(aq). At a particular temperature the rate of disappearance of S2O2−8 varies with reactant concentrations in the following manner: Experiment S2O2−8(M) I−(M) Initial Rate (M/s) 1 0.018 0.036 2.6×10−6 2 0.027 0.036 3.9×10−6 3 0.036 0.054 7.8×10−6 4 0.050 0.072 1.4×10−5 What is the rate of disappearance of I− when [S2O2−8]= 1.8×10−2 M and [I−]= 5.0×10−2 M ?
The rate law for the reaction is rate = k[S2O8²⁻], and the rate of disappearance of I- can be determined using the stoichiometric coefficients.
Explanation:The rate law for the reaction of peroxydisulfate ion (S2O8²⁻) with iodide ion (I⁻) can be determined by using the initial rates data. In the given data, we can see that the initial rate varies linearly with the concentration of S2O8²⁻ and is independent of the concentration of I⁻. This means that the rate law for this reaction is rate = k[S2O8²⁻].
Using the given rate law, we can determine the rate of disappearance of I⁻ when [S2O8²⁻] = 1.8×10⁻²M and [I⁻] = 5.0×10⁻²M. Since the stoichiometric coefficient of I⁻ in the balanced chemical equation is 3, the rate of disappearance of I⁻ is 3 times the rate of appearance of S2O8²⁻. Therefore, the rate of disappearance of I⁻ is (3)(1.4×10⁻⁵M/s) = 4.2×10⁻⁵M/s.
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You just measured a metal cylinder and obtained the following information: mass - 3.543 g diameter -0.53 cm height = 4.40 cm. 265 • Determine the volume (V). (V=nrºh, where r = radius, h = height, T - 3.14) V= (3.14 36205² 4.402.17cm? • Determine density of the cylinder D. 3. SMS-365923 197
The density of the cylinder would be 3.652 gram/ cm³
.
What is density?It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.
As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,
mass of the cylinder = 3.543 grams
the volume of the cylinder = πr²h
= 3.14 ×.265²×4.4
=0.97 cm³
By using the above formula for density
ρ = mass of the cylinder/volume of the cylinder
= 3.543 grams/0.97 cm³
=3.652 grams/ cm³
Thus,the density of the cylinder would be 3.652 grams/ cm³.
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To find the volume and density of a metal cylinder with a mass of 3.543 g, diameter of 0.53 cm, and height of 4.40 cm, we first calculate the volume as 3.08 cm³ using the formula V = πr²h. Then, we determine the density to be 1.15 g/cm³ by dividing the mass by the volume.
You just measured a metal cylinder and obtained the following information: mass - 3.543 g, diameter - 0.53 cm, height = 4.40 cm. To determine the volume and density of the cylinder, follow these steps:
Determining the Volume (V)
First, understand that the volume (V) of a cylinder can be calculated using the formula V = πr²h, where π (pi) is approximately 3.14, r is the radius of the cylinder (half of the diameter), and h is the height of the cylinder. Given a diameter of 0.53 cm, the radius is 0.53 cm / 2 = 0.265 cm. Substituting these values into the formula gives:
V = 3.14 * (0.265 cm)² * 4.40 cm = 3.08 cm³
Determining the Density (D)
Density can be found using the formula D = m / V, where m is the mass and V is the volume. Using the mass of 3.543 g and the volume of 3.08 cm³:
D = 3.543 g / 3.08 cm³ = 1.15 g/cm³
Tracy is studying for her chemistry exam and she reads that a nitric acid molecule is made of one hydrogen atom, one nitrogen atom and 3 oxygen atoms. The chemical formula for nitric acid is a. HNO3 Oьно OCHN O d.HO3 Oe. HN30
Answer:
a. HNO₃
Explanation:
Nitric acid -
The chemical formula for nitric acid is - HNO₃ , with molecular mass of 63 g /mol , having one hydrogen atom, one nitrogen atom and 3 oxygen atoms .
Nirtic acid is a colorless solution , but the left over solution turns yellow as it get decomposed to nitrogen oxide and water .
The commercially used nitric acid is 68 % in water .
and more than 86% nitric acid is fuming nitric acid .
The reagent nitric acid is used a source of nitration .
the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K298 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00300 M[glyceraldehyde-3-phosphate]=0.00300 M .
Answer:
ΔG = 16.218 KJ/mol
Explanation:
dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphateΔG = ΔG° - RT Ln Q∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol
∴ R = 8.314 J/K.mol
∴ T = 298 K
∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]
⇒ Q = 0.00300 / 0.100 = 0.03
⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))
⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )
The vapour pressure of water at 20 C is 2.34 kPa. Given that the heat of vaporisation is 2537.4 kJ/kg, use the Clausius-Clapeyron equation to give the vapour pressure at 40 C.
The Clausius-Clapeyron equation is used to calculate the vapor pressure of water at 40°C given its vapor pressure at 20°C and heat of vaporisation. By rearranging the equation and inserting the known values, the vapor pressure at 40°C can be found, which will be higher than that at 20°C.
Explanation:The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance to its heat of vaporisation. To calculate the vapor pressure of water at 40°C, given the vapor pressure at 20°C (2.34 kPa) and the heat of vaporisation (2537.4 kJ/kg), we can rearrange the equation to solve for the new vapor pressure:
Clausius-Clapeyron Equation: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where P1 is the initial vapor pressure, P2 is the final vapor pressure, ΔHvap is the molar heat of vaporisation, R is the universal gas constant in J/(mol·K), T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.
The calculation involves converting both temperatures from Celsius to Kelvin and plugging the values into the equation to solve for P2. After using this equation, we find that the vapor pressure at 40°C is significantly higher than at 20°C.
Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base reaction HCI + H20 → H30+ + Cl- a) Is this a strong acid? b) Clearly label the acid, base, conjugate acid and conjugate base. (5 points)
Answer: Yes, [tex]HCl[/tex] is a strong acid.
acid = [tex]HCl[/tex] , conjugate base = [tex]Cl^-[/tex] , base = [tex]H_2O[/tex], conjugate acid = [tex]H_3O^+[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
Yes [tex]HCl[/tex] is a strong acid as it completely dissociates in water to give [tex]H^+[/tex] ions.
[tex]HCl\rightarrow H^++Cl^-[/tex]
For the given chemical equation:
[tex]HCl+H_2O\rightarrow H_3O^-+Cl^-[/tex]
Here, [tex]HCl[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]Cl^-[/tex] which is a conjugate base.
And, [tex]H_2O[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_3O^+[/tex] which is a conjugate acid.
Thus acid = [tex]HCl[/tex]
conjugate base = [tex]Cl^-[/tex]
base = [tex]H_2O[/tex]
conjugate acid = [tex]H_3O^+[/tex].
Calculate the number of atoms of bromine in 1.37 g of bromine. Give your answer in scientific notation. Enter your answer in the provided box. x 10 (select)atoms Br
Answer: The number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of bromine = 1.37 g
Molar mass of bromine = 79.904 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of bromine}=\frac{1.37g}{79.904g/mol}=0.0171mol[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 0.0171 moles of bromine will contain = [tex]0.0171\times 6.022\times 10^{23}=1.03\times 10^{22}[/tex] number of bromine atoms.
Hence, the number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]
Final answer:
To find the number of bromine atoms in 1.37 g of bromine, calculate the atomic mass using isotope information, convert grams to moles, and then use Avogadro's number, yielding approximately 1.03 × 10²² bromine atoms.
Explanation:
To calculate the number of bromine atoms in 1.37 g of bromine, we first need to determine the molar mass of bromine. The atomic mass of bromine is calculated based on the isotopic composition, considering Bromine has two isotopes, 79Br and 81Br with masses of 78.9183 amu and 80.9163 amu respectively, and relative abundances of 50.69% and 49.31%. Using these values, we calculate the average atomic mass of bromine:
Atomic mass = (0.5069 × 78.9183 amu) + (0.4931 × 80.9163 amu) = 79.904 amu.
We then convert grams of bromine to moles by dividing by the molar mass:
Number of moles = 1.37 g ÷ 79.904 g/mol = 0.01715 moles.
Using Avogadro's number, 6.022 × 10²³ atoms/mol, we multiply the number of moles of bromine by this constant to find the number of atoms:
Number of atoms = 0.01715 moles × 6.022 × 10²³ atoms/mol = 1.03 × 10²² atoms.
Thus, there are 1.03 × 10²² atoms of bromine in 1.37 g of bromine. Remember, when giving your answer in scientific notation, it is important to express it in the form of a number between 1 and 10 multiplied by a power of 10.
If a microwave oven emits 1,100 J of energy every second, how long will it take for the microwave to emit 4650 J of energy? Express your answer in seconds using the correct number of significant figures. Do not enter your answer using scientific notation.
It will take the microwave approximately 4.2 seconds to emit a total of 4650 J of energy, calculated by dividing the total energy by the power it emits per second, using two significant figures.
To calculate how long it will take for a microwave emitting 1,100 J of energy every second to emit a total of 4650 J, we can use the formula:
Time (s) = Total Energy (J) / Power (J/s).
Substituting the given values:
Time (s) = 4650 J / 1,100 J/s = 4.22727272727... seconds
Since the microwave emits energy at a rate of 1,100 J every second, we want to express the answer with the same number of significant figures as the least precise measurement, which is two significant figures. Therefore, the time should be rounded to 4.2 seconds.
How many grams of precipitate can be formed in the reaction between 6.00 mL of 0.10 M silver nitrate and 5.00 mL of 0.15 M potassium chloride? Hint: Start with a balanced chemical equation. Do not enter units with your answer.
Answer:
0.09 g of precipitate can be formed.
Explanation:
The chemical equation can be written as follows:
AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺
From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.
The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:
n° of moles of silver nitrate = concentration * volume
n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃
n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl
Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.
1.5 x 10⁻⁴ mol KCl is in excess.
The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.
A process stream contains 450 mol/s benzene and 375 mol/s toluene. Calculate the mole fraction of benzene in the process stream.
Answer:
0.5455
Explanation:
The moles of benzene in the process stream in 1 sec = 450 moles
The moles of toluene in the process stream in 1 sec = 375 moles
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]
Applying values as:
[tex]Mole\ fraction\ of\ benzene=\frac {450}{450+375}[/tex]
Mole fraction of benzene in the process stream = 0.5455
If you have 0.08157 moles of Al(CIO4) 3... a) How many moles of Al ions do you have? b) How many moles of CIO4 ions do you have? c) How many moles of oxygen atoms do you have? 3+
Answer: a) 0.08157 moles of [tex]Al^{3+}[/tex]
b) 0.24471 moles of [tex]ClO_4^{-}[/tex]
c) 0.97884 moles of oxygen atoms
Explanation:-
The dissociation of the given compound is shown by the balanced equation:
[tex]Al(ClO_4)_3\rightarrow Al^{3+}+3ClO_4^{-}[/tex]
According to stoichiometry:
a) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 1 mole of [tex]Al^{3+}[/tex]
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{1}{1}\times 0.08157=0.08157moles[/tex] of [tex]Al^{3+}[/tex]
b) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 3 moles of [tex]ClO_4^{-}[/tex]
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{3}{1}\times 0.08157=0.24471moles[/tex] of [tex]ClO_4^{-}[/tex]
c) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 12 moles of oxygen atoms
Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{12}{1}\times 0.08157=0.97884moles[/tex] of oxygen atoms
Calculate the amount of CO2 (in kg) released when 1 kg coal is burnt. Assume that carbon content of the coal is 50% by mass. 4. Calculate the CO2 production in kg/MJ if a coal fired power plant (efficiency - 30%) is used to produce electricity. Assume energy density of coal - 24 MJ/kg and assume the coal is
Explanation:
(a) Here, the assumption is that complete combustion of carbon is taking place. This means that there is no CO formation
Reaction equation when carbon is burning is as follows.
[tex]C + O_{2} \rightarrow CO_{2}[/tex]
Molar mass of C = 12 kg/kmol
Molar mass of [tex]CO_{2}[/tex] = 44 kg/kmol
Hence, according to the stoichiometry,
1 Kmol of C reacted = 1 kmol of [tex]CO_{2}[/tex] produced
Therefore, 12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced
It is given that coal contains 50 mass % Carbon
So, 1 kg of coal contains (0.50 × 1)kg Carbon
Carbon in 1 kg Coal = 1 × 0.5 = 0.5 kg
As per the stoichiometry,
12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced
0.5 kg of C reacted = x kg [tex]CO_{2}[/tex] produced
Therefore, value of x can be calculated as follows.
x = [tex]\frac{\text{44 kg of CO_{2} produced} \times \text{0.5 kg of C reacted}}{\text{12 kg of C reacted}}[/tex]
= 1.83 kg
This means that amount of [tex]CO_{2}[/tex] released is 1.83 kg.
(b) It is assumed that coal contains 50 mass % carbon and 1 kg of coal burnt.
Since, it is given that energy density of coal is 24 Mj and efficiency of the power plant is 30%.
After burning 1 kg of coal amount of energy released = 24 Mj
Amount of energy converted to electricity = [tex]24 Mj \times 0.3[/tex] = 7.2 Mj
It is calculated that amount of [tex]CO_{2}[/tex] released per 1 kg of coal = 1.83 kg
Therefore, calculate the amount of [tex]CO_{2}[/tex] released in kg/Mj as follows.
amount of [tex]CO_{2}[/tex] released in kg/Mj = [tex]\frac{\text{Amount of CO_{2} released in kg}}{\text{amount of energy converted to electricity}}[/tex]
= [tex]\frac{1.83 kg}{7.2 mJ}[/tex]
= 0.2541 kg/Mj
Hence, the production of [tex]CO_{2}[/tex] in kg/MJ is 0.2541 kg/Mj.
To calculate the amount of CO2 produced from 1 kg of coal, first determine the mass of carbon, apply the conversion for the mass of CO2 produced per mass of carbon, and use the plant's efficiency and the coal's energy density to determine CO2 production per MJ.
Explanation:The question involves consulting basic chemical stoichiometry and energy conversion to solve problems related to carbon dioxide emissions and energy efficiency in the context of coal combustion. Knowing the percentage of carbon in coal, we can determine the mass of carbon dioxide produced from burning coal. Additionally, we'll use the energy density of coal to calculate CO2 production per unit energy output of a coal-fired power plant.
To find the mass of CO2 produced from burning 1 kg of coal with 50% carbon content, we first calculate the mass of carbon that would burn:
Mass of carbon in 1 kg of coal = 1 kg * 50% = 0.5 kg.The molecular weight of carbon is approximately 12 g/mol, and for CO2 it is approximately 44 g/mol. This implies that every 12 kg of carbon produces 44 kg of CO2. For 0.5 kg (or 500 g) of carbon, the mass of CO2 produced would be:
(44 kg CO2 / 12 kg C) * 0.5 kg C = 1.833 kg CO2.For the second part, to calculate the CO2 production in kg/MJ for a coal-fired power plant with 30% efficiency:
Energy content per kg coal = 24 MJ/kg.Effective energy per kg coal (due to 30% efficiency) = 24 MJ/kg * 30% = 7.2 MJ/kg.Thus, for 1 kg of coal combusted, 1.833 kg of CO2 is released. This amount of CO2 is produced for every 7.2 MJ of energy output.CO2 production per MJ = 1.833 kg CO2 / 7.2 MJ = 0.2546 kg CO2/MJ.
A decomposition reaction has a rate constant of 0.0012 yr^-1
(a) What is the half-life of the reaction?
(b) How long does it take for [reactant] to reach 15% of its original value?
Answer:
A) 577.6 years
B) 1580.9 years
Explanation:
You can know the order of a reaction given the units of the rate constant:
For a 0 order reaction are M/s=(mol/L*s)For a 1 order reaction are 1/sFor a 2 order reaction are 1/M*s=(L/mol*s)Then we know that this is a first order reaction because years is a unit of time as well as seconds. The half life of a first order reaction is given by:
[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]
Here you can solve for the half life:
[tex]t_{1/2}=\frac{ln(2)}{0.0012 years^-1}=577.6 years[/tex]
Now for the rate law for a first order reaction is:
[tex][A]=[A_{o} ]*e^{-kt}[/tex]
Then, if you want to know how long does it take to reach a certain value you solve for time:
[tex]t=-ln([A]/[A]_{o}) *1/k[/tex]
The 15% of its original value is [tex]0.15*[A_{o}][/tex]
You solve for time:
[tex]t=-ln(0.15) *1/0.0012 years^-1=1580.9 years[/tex]
Hope it helps!
In the Decomposition reaction, a reactant gets split into two or more products. The half-life of the reaction is 577.6 years and the time taken to reach 15% of the original is 1580.9 years.
What is half-life?Half-life is the time taken by the radioactive isotope to decay one-half of its total amount.
From the units of the rate of the reaction, the order can be estimated as, for a first-order reaction unit is 1/s.
The half-life of the reaction can be given as,
[tex]\rm t\dfrac{1}{2} = \dfrac{\rm ln (2)}{k}[/tex]
The rate constant (k) for the decomposition reaction is 0.0012 per year.
Substituting values in the above equation:
[tex]\begin{aligned} \rm t\dfrac{1}{2} &= \dfrac{\rm ln (2)}{0.0012}\\\\&= 577.6\;\rm years\end{aligned}[/tex]
The rate law of the first-order reaction is given as,
[tex]\begin{aligned}\rm [A] = [A_{o}] \times e^{-kt}\\\\\rm t = -ln ([A][A_{o}]) \times \dfrac{1}{k}\end{aligned}[/tex]
Solving for time (t):
[tex]\begin{aligned}\rm t &= \rm -ln (0.15) \times \dfrac{1}{0.0012}\\\\& = 1580.9\;\rm years\end{aligned}[/tex]
Therefore, the half-life is 577.6 years and the time taken is 1580.9 years.
Learn more about half-life here:
https://brainly.com/question/18802932
If I want to accelerate a mass of 3 kg at 5 m/s2 then how much force should I apply?
Answer: The force that must be applied is 15 N.
Explanation:
Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.
Mathematically,
[tex]F=m\times a[/tex]
where,
F = force exerted = ?
m = mass of the object = 3 kg
a = acceleration of the object = [tex]5m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=3kg\times 5m/s^2=15N[/tex]
Hence, the force that must be applied is 15 N.
Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.
Answer:
You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate
[tex]3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g[/tex]
Explanation:
First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.
By increasing the length of heat exchanger, the effectiveness and cold stream output temperature of the heat exchanger
a- Decreases
b- Increases
c- Does not change
Answer:
Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.
Explanation:
We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.
Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.
In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.
If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.
Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.
When a_ ( sublimes, a converts to a_ (ii) This phase change is_ because energy is_(iv) when the intermolecular forces_ [M]__ (i) (ii) (iii) (iv) (v) a. solid Liquid endothermic absorbed form b. solid Gas endothermic absorbed break c. solid Liquid exothermic released break solid Gas exothermic released form e gas Solid exothermic released form d
Answer: Option (b) is the correct answer.
Explanation:
Sublimation is defined as the process in which a solid substance changes directly into gaseous phase without undergoing into liquid phase.
For example, when naphthalene balls are kept in a trunk or cupboard then after some days or months it changes into gas. Hence, they become consumed.
Also during this process energy is being absorbed by the solid substance because for breaking bonds energy is always absorbed by a substance. Only then it can change into liquid or vapor state.
So, a chemical process in which energy is being absorbed by a reaction is known as endothermic reaction. And, if heat energy is being released by the reaction then it is known as an exothermic reaction.
Therefore, we can conclude that when a solid sublimes, a solid converts to a gas. This phase change is endothermic because energy is absorbed when the intermolecular forces break.
A sailor circumnavigated the earth and covered 4.264,000 meters. Express this number in standard scientific notation A) 4.264 . 10-6 m B ) 4.264 . 10 m C) 4.264 . 106m D) 4.264 . 10-7m
Answer:
[tex]4.264*10^{6}[/tex]
Explanation:
Hello !
The correct answer is option C) [tex]4.264*10^{6}[/tex]
To write in scientific notation, numbers between 0 and 9 are used before the point and the rest of the numbers afterwards.
Then we count how many places we ran the point. If you run to the right, the exponent of 10 is positive and if you run to the left, it is negative.
This way it is [tex]4.264*10^{6}[/tex]