Draw the products for the proton transfer reaction between sodium hydride and ethanol

Answer:

Silver (Ag) is more reactive than platinum (Pt) :) hope this helps!!!

Explanation:

Answer:

0.0342 mol

Explanation:

The molar mass of boron is 10.81 g/mol, that is, 1 mole of boron (6.02 × 10²³ molecules of boron) has a mass of 10.81 grams. This is the ratio that we will use to find the number of moles in 3.70 × 10⁻¹ g, using a conversion fraction.

3.70 × 10⁻¹ g B × (1 mol B / 10.81 g B) = 0.0342 mol B

To find the number of moles in Boron, you divide the given mass by the molar mass of Boron. Given the mass of Boron as 3.70 x 10^-1 g and the molar mass as 10.8 g/mol, the result is approximately 0.034 moles.

To find the number of moles in a sample, we use the formula: Moles = mass / molar mass. We already know the mass of boron, which is 3.70 x 10^-1 g. The average atomic mass of boron, considering its isotopes, is approximately 10.8 amu. Converting this into grams gives you 10.8 g/mol (since 1 amu = 1 g/mol).

Therefore, the number of moles of boron is calculated to be Moles = 3.70 x 10^-1 / 10.8 = approximately 0.034 moles.

Note that an atomic mass unit (amu) is basically the mass of one atom, on a scale where the carbon-12 atom is exactly 12.0 amu. However, no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, considering the isotopes.

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The equilibrium constant is the proportion of the equilibrium concentration of the product to the reactants. The moles of CO present in a flask is 1.89 moles.

Moles are the product of the molar concentration and the volume of the solution. It is given in mol.

The balanced chemical reaction can be shown as:

[tex]\rm 2CO(g) + O_{2}(g) \rightarrow 2CO_{2}(g)[/tex]

The equilibrium constant is given as,

[tex]\rm K_{c} = \dfrac{[CO_{2}]^{2}}{[CO_{2}]^{2} [O_{2}]}[/tex]

The concentration of carbon dioxide is calculated as:

[tex]\dfrac{0.4 \;\rm mol}{3\;\rm L} = 0.13[/tex]

The concentration of oxygen is calculated as:

[tex]\dfrac{0.1 \;\rm mol}{3\;\rm L} = 0.03[/tex]

Substituting values in the formula of the equilibrium constant:

[tex]\begin{aligned}{1.4 \times 10^{2} &= \rm \dfrac{0.13^{2}}{ [CO]^{2} \times 0.03}\\\\\rm [CO] &= \sqrt{0.004} \\\\&= 0.63 \end{aligned}[/tex]

The moles of carbon monoxide will be  [tex]0.63 \times 3 = 1.89\;\rm moles.[/tex]

Therefore, 1.89 moles of carbon monoxide are present in the flask.

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The correct answer is that there are 0.200 moles of CO present in the flask at equilibrium.

To solve this problem, we need to consider the chemical equilibrium of the reaction involved, which is the decomposition of carbon dioxide (CO2) into carbon monoxide (CO) and oxygen (O2). The balanced chemical equation for this reaction is:

[tex]\[ \text{CO}_2(g) \rightleftharpoons \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \][/tex]

According to the stoichiometry of the balanced equation, for every mole of CO2 that decomposes, one mole of CO is produced and half a mole of O2 is produced.

Given that we have 0.400 moles of CO2 and 0.100 moles of O2 in the flask, we can use the stoichiometry of the reaction to find out how many moles of CO are present. Since the reaction produces one mole of CO for every mole of CO2 that decomposes, we can calculate the moles of CO produced by the decomposition of CO2 as follows:

[tex]\[ \text{Moles of CO produced} = \text{Moles of CO2 decomposed} \][/tex]

However, we also know that the reaction produces half a mole of O2 for every mole of CO2 that decomposes. Therefore, the moles of O2 produced can be calculated as:

[tex]\[ \text{Moles of O2 produced} = \frac{1}{2} \times \text{Moles of CO2 decomposed} \][/tex]

Since we have 0.100 moles of O2, we can set up the equation:

[tex]\[ 0.100 = \frac{1}{2} \times \text{Moles of CO2 decomposed} \][/tex]

Solving for the moles of CO2 decomposed, we get:

[tex]\[ \text{Moles of CO2 decomposed} = 0.100 \times 2 = 0.200 \][/tex]

This means that 0.200 moles of CO2 have decomposed to produce 0.200 moles of CO and 0.100 moles of O2. Therefore, the moles of CO present in the flask at equilibrium is 0.200 moles.

The compound Fe(ClO4)2 in water would dissociate into its ions, which are Fe²+ and 2ClO4¯. A specific net ionic equation cannot be given without knowledge of the reactants.

The net ionic equation would be the result of considering all the ions in the reaction of Fe(ClO4)2, which is iron(II) perchlorate, with all possible reactants. However, without knowing these reactants, we can't provide a specific net ionic equation. Generally, important concepts here include understanding how to balance equations and predict solubility based on common rules. In the case of Fe(ClO4)2 in water, this compound is highly soluble and would dissociate into its ions in water. Generally, you would expect it to dissociate into Fe²+ and 2ClO4¯ ions in solution.

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Final answer:

The correct statement for a chemical reaction with an equilibrium constant of 0.213 is that there are more reactants than products at equilibrium.

Explanation:

Given that the equilibrium constant is 0.213 for a chemical reaction, we can make certain determinations about the state of the reaction at equilibrium. When the equilibrium constant (K) is less than 1, this generally means that the ratio of products to reactants at equilibrium is small and the reaction system favors the reactants. Therefore, the correct statement is that there are more reactants than products at equilibrium.

It is important to understand that when equilibrium is reached, the reaction has not stopped; rather, it is a dynamic state where the forward and reverse reactions continue at the same rate, maintaining constant concentrations of reactants and products. The idea that the reaction continues until no reactant remains is incorrect because a reaction at equilibrium does not favor complete conversion to products unless the equilibrium constant is significantly greater than 1.

Answer: C) develop a way to adjust its density so that it can be above and below 1.0 gram per cubic centimeter.

Explanation: Submarine is a kind of special ship that can float above and below water. We know that an object can only float over water when its density is less than the density of water and it would sink if its density is greater than the density of water.

A submarine has tanks in which water is filled if we want to sink it and the water from the tanks is taken out with the help of water pumps to rise the submarine. These water tanks in the submarine helps to adjust its density.

The over all density is increased by filling the water tanks to sink the submarine and the density is decreased by removing the water from the tanks to rise the submarine.

So, option third or C is correct.

Set the crucible's lid slightly off-center to allow air to enter while keeping the magnesium oxide from escaping.

A crucible is a cup-shaped piece of laboratory equipment used to keep chemical compounds contained while they are heated to extremely high temperatures.

Crucibles come in a variety of sizes and are usually packaged with a crucible cover (or lid).

Keep the lid on the crucible while cooling to prevent moisture from the atmosphere from interacting with the anhydrous salt, especially if the lab is humid. As a result, the mass of water will be too low.

The most important apparatus because it will be used to obtain the final precipitate, which will tell us how much salt is in the solution.

The lid is used to cover the crucible so that the heated precipitates do not oxidize when they come into contact with air.

Thus, it is important that the lid should be kept correctly.

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Answer:

Electrons will be gained

oxides and carbonates, grad point..

Answer: The element needs to react with other element by gaining an electron to complete its valence electron.

Explanation: The element which exhibit 7 valence electrons are halogens. They readily react with other element, for example: Hydrogen and Sodium, in order to gain an electron to complete their outermost shell.

Answer:

Mass is a measure of the quantity of matter in an object

Explanation:

Answer:

The mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.

Explanation:

Hello,

In case, no matter the carried out experiment, the law of conservation of mass always leads the same: the mass can not be neither created nor destroyed but transformed by chemical reactions or physical transformations in an isolated recipient.

In such a way, we must consider that any system closed to every form of transport of matter, will show a no change in its mass as time goes by, since the system's mass cannot change neither by additions nor withdrawals. Therefore, the quantity of mass is conserved over time.

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Answer is: the final pressure of the gas is closest to 3,17 atm.

p₁ = 3,2 atm.
T₁ = 310 K.
V₁ = 6 L.
p₂ = ?
T₂ = 410K.
V₂ = 8,0 L.
Use combined gas law - the volume of amount of gas is proportional to the ratio of its Kelvin temperature and its pressure. 
p₁V₁/T₁ = p₂V₂/T₂.
3,2 atm · 6,0 L ÷ 310 K  = p₂ · 8,0 L ÷ 410 K.
0,0619 = 0,0195p₂.
p₂ = 3,17 atm.

It will require  for 20 % of U-238 atoms in a sample of U-238 to decay.

Further Explanation:

Radioactive decay involves stabilization of unstable atomic nucleus and is accompanied by the release of energy. This emission of energy can be in form of different particles like alpha, beta and gamma particles.

Half-life is time period in which half of the radioactive species is consumed. It is denoted by .

The expression for half-life is given as follows:

[tex]\lambda = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex]                    …… (1)

Where,

[tex]{t_{{\text{1/2}}}}[/tex] is half-life period

[tex]\lambda[/tex] is the decay constant.

The half-life period for decay of U-238 is [tex]4.5 \times {10^9}{\text{ yrs}}[/tex].

Substitute [tex]4.5 \times {10^9}{\text{ yrs}}[/tex] for [tex]{t_{{\text{1/2}}}}[/tex]  in equation (1).

[tex]\begin{aligned}\lambda&= \dfrac{{0.693}}{{4.5 \times {{10}^9}{\text{ yrs}}}} \\&= 1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}} \\\end{galigned}[/tex]  

Since it is radioactive decay, it is first-order reaction. Therefore the expression for rate of decay of U-238is given as follows:

[tex]\lambda = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)[/tex]

                                                …… (2)

Where,

[tex]\lambda[/tex] is the decay or rate constant.

t is the time taken for decay process.

a is the initial amount of sample.

x is the amount of sample that has been decayed.

Rearrange equation (2) to calculate t.

[tex]t = \dfrac{{2.303}}{\lambda }\log \left( {\dfrac{a}{{a - x}}} \right)[/tex]                                                                                          …… (3)

Consider 100 g to be initial amount of U-238. Since 20 % of it is decayed in radioactive process, 20 g of U-238 is decayed and therefore 80 g of the sample is left behind.

Substitute 100 g for a, 80 g for (a–x) and [tex]1.54 \times {10^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}[/tex] for [tex]\lambda[/tex] in equation (3).

[tex]\begin{aligned}t &= \dfrac{{2.303}}{{1.54 \times {{10}^{ - 10}}{\text{ yr}}{{\text{s}}^{ - 1}}}}\log \left( {\dfrac{{100{\text{ g}}}}{{80{\text{ g}}}}} \right)\\&= 1.449 \times {10^9}{\text{ yrs}}\\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, t, a, x, a – x, 1.449*10^9 yrs, U-238, decay constant, radioactivity, half-life period.

The new pH of given solutionis [tex]\boxed{3.53}[/tex].

Further Explanation:

The aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is termed as buffer solution. These solutions offer strong resistance to any change in their pH on addition of small quantity of strong acid or base.

Henderson-Hasselbalch equation:

This equation helps in determining the pH of buffer solution. Its mathematical form is given as follows:

[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}[/tex]                                            …… (1)

Here,

[tex]\left[ {{{\text{A}}^ - }} \right][/tex] is concentration of conjugate base.

[HA] is concentration of acid.

Given mixture is a buffer solution of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] and [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex]. Therefore Henderson-Hasselbalch equation becomes as follows:

[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right]}}{{\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right]}}[/tex]                                                  …… (2)

Initial moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:

[tex]\begin{aligned}{\text{Moles of HN}}{{\text{O}}_2} &= \left( {1.2{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 1.2{\text{ mol}} \\\end{aligned}[/tex]  

Initial moles of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex]  can be calculated as follows:

[tex]\begin{aligned}{\text{Moles of NaN}}{{\text{O}}_{\text{2}}} &= \left( {0.8{\text{ M}}} \right)\left( {{\text{1 L}}} \right)\\&= 0.8{\text{ mol}} \\\end{aligned}[/tex]  

Moles of NaOH can be calculated as follows:

[tex]\begin{aligned}{\text{Moles of NaOH}} &= \left( {0.5{\text{ M}}} \right)\left( {{\text{400 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.2{\text{ mol}} \\\end{aligned}[/tex]  

When addition of 0.2 moles of NaOH is done to the buffer solution, 0.2 moles of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] is neutralized while the same amount of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] is formed. Since volumes are additive, total volume can be calculated as follows:

[tex]\begin{aligned}{\text{Total volume of solution}} &= \left( {1 + \left( {400{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)} \right){\text{ L}} \\ &= {\text{1}}{\text{.4 L}} \\\end{aligned}[/tex]

Therefore concentration of [tex]{\text{HN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:

 [tex]\begin{aligned}\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 - 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}}\\&= 0.714{\text{ M}} \\\end{aligned}[/tex]

Therefore concentration of [tex]{\text{NaN}}{{\text{O}}_{\text{2}}}[/tex] can be calculated as follows:

[tex]\begin{aligned}\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right] &= \frac{{\left( {1.2 + 0.2} \right){\text{ mol}}}}{{1.4{\text{ L}}}} \\ &= 1{\text{ M}} \\\end{aligned}[/tex]  

Substitute 0.714 M for [tex]\left[ {{\text{HN}}{{\text{O}}_{\text{2}}}} \right][/tex], 1 M for [tex]\left[ {{\text{NaN}}{{\text{O}}_{\text{2}}}} \right][/tex] and 3.4 for [tex]{\text{p}}{K_{\text{a}}}[/tex] in equation (2).

[tex]\begin{aligned} {\text{pH}} &= 3.4 + {\text{log}}\left( {\frac{{{\text{1 M}}}}{{0.714{\text{ M}}}}} \right) \\&= 3.54 \\\end{aligned}[/tex]  

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acid, base and salts

Keywords: pH, buffer, pKa, NaNO2, HNO2, 3.4, 1 M, 0.714 M, concentration, total volume of solution, 1.2 mol, 0.8 mol, 0.2 mol, 3.54.

To prepare a 300.0 mL of a 1.50 M KBr solution, you need 53.55 grams of KBr, calculated by multiplying the required moles (0.450 moles) by the molar mass of KBr (119.00 g/mol).

To find the mass of KBr needed to make a 300.0 mL (0.300 L) of a 1.50 M solution, we apply the formula:

Molarity (M) = moles of solute / liters of solution

First, calculate the moles of KBr required:

Next, convert moles to grams using the molar mass of KBr (119.00 g/mol):

Therefore, you would need 53.55 grams of KBr to make a 300.0 mL of a 1.50 M KBr solution.

Answer: The molality of solution is 1.94 m.

Explanation:

We are given:

Mass of sulfuric acid = 17.75 grams

Volume of solution = 100 mL

Density of solution = 1.1094 g/mL

To calculate the mass of solution, we use the equation:

[tex]Density=\frac{Mass}{Volume}[/tex]

Putting values in above equation, we get:

[tex]1.1094g/mL=\frac{\text{Mass of solution}}{100mL}[/tex]

[tex]\text{Mass of solution}=110.94g[/tex]

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent = 110.94 - 17.75 = 93.19 g

To calculate the molality of solution, we use the equation:

[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](H_2SO_4)[/tex] = 17.75 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](H_2SO_4)[/tex] = 98 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent = 93.19 g

Putting values in above equation, we get:

[tex]\text{Molality of }H_2SO_4=\frac{17.75\times 1000}{98\times 93.19}[/tex]

[tex]\text{Molality of }H_2SO_4=1.94m[/tex]

Hence, the molality of solution is 1.94 m.

To calculate the molality of the sulfuric acid solution, we convert the volume to mass, find the number of moles of sulfuric acid, and divide by the mass of the solvent in kilograms.

To calculate the molality of a solution, we need to find the number of moles of solute and the mass of the solvent.

In this case, we are given the mass of sulfuric acid (17.75 g) and the volume of the solution (100.0 ml). First, we need to convert the volume to mass by multiplying it by the density of the solution (1.1094 g/ml). 100.0 ml x 1.1094 g/ml = 110.94 g.

To find the number of moles of sulfuric acid, we divide the mass by the molar mass of sulfuric acid (98.09 g/mol). 17.75 g / 98.09 g/mol ≈ 0.1808 mol.

Finally, we can calculate the molality by dividing the moles of solute by the mass of the solvent in kilograms. 0.1808 mol / 0.11094 kg = 1.63 mol/kg.

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When the concentration of b is doubled in a chemical reaction that is zero order in a, one-half order in b, and second order in c, the reaction rate will increase by a factor of √2 (approximately 1.414), with other reactant concentrations held constant.

The question you're asking relates to the rate of a chemical reaction and how it changes with varying concentrations of reactants. Specifically, there is a reaction where the rate is zero order in a, one-half order in b, and second order in c. According to the given reaction orders, the rate expression would be:

Rate = k [a]0[b]1/2[c]2

Since the reaction is zero order in a, changing the concentration of a does not affect the rate. However, since it is one-half order in b, if the concentration of b is doubled, the rate will increase by a factor of the square root of 2. This is because:

New Rate = k [a]0(2[b])1/2[c]2 = k [a]0[b]1/2 × √2 [c]2
= Rate × √2

Therefore, when the concentration of b is doubled, and a and c remain constant, the reaction rate will increase by a factor of √2 (approximately 1.414).

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When the size of a star increases, the brighter it gets.

The color of the star is determined by the temperature.

If the temperature of the star is lower, that means the star is an orange/red color.

If the star's temperature is higher, that means the star is an blue/white color.