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4 years ago

Type in he correct values to correctly represent the valence electron configuration of oxygen

Chemistry

1 answer:

Aliun [14]4 years ago

7 0

2s2 2p4 is the valence electron configuration for oxygen

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Suppose 2.19g of barium acetate is dissolved of 15oml barium of a 0.10M acetate is dissolved in of a aqueous solution of sodium

Jobisdone [24]

The question is incomplete, here is the complete question:

Suppose 2.19 g of barium acetate is dissolved in 150 mL of a 0.10M of aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer: The final molarity of acetate ion in the solution is 0.12 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

For Sodium chromate:

Molarity of sodium chromate solution = 0.10 M

Volume of solution = 150 mL

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of sodium chromate}\times 1000}{150}\\\\\text{Moles of sodium chromate}=\frac{(0.10\times 150)}{1000}=0.015mol$

For barium acetate:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of barium acetate = 2.19 g

Molar mass of barium acetate = 255.43 g/mol

Putting values in above equation, we get:

$\text{Moles of barium acetate}=\frac{2.19g}{255.43g/mol}=0.0086mol$

The chemical equation for the reaction of barium acetate and sodium chromate follows:

$Ba(CH_3CO_2)_2+Na_2CrO_4\rightarrow BaCrO_4(s)+2Na^+(aq.)+2CH_3CO_2^-(aq.)$

By stoichiometry of the reaction:

1 mole of barium acetate reacts with 1 mole of sodium chromate

So, 0.0086 moles of barium acetate will react with = $\frac{1}{1}\times 0.0086mol$ of sodium chromate

As, given amount of sodium chromate is more than the required amount. So, it is considered as an excess reagent.

Thus, barium acetate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of barium acetate produces 2 moles of acetate ions

So, 0.0086 moles of barium acetate will produce = $\frac{2}{1}\times 0.0086=0.0172mol$ of acetate ion

Now, calculating the molarity of acetate ions in the solution by using equation 1:

Moles of acetate ion = 0.0172 moles

Volume of solution = 150 mL

Putting values in equation 1, we get:

$\text{Molarity of acetate ions}=\frac{0.0172\times 1000}{150}\\\\\text{Molarity of acetate ions}=0.12M$

Hence, the final molarity of acetate ion in the solution is 0.12 M

5 0

4 years ago

The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the

deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is $0.3115 \ \text{sec}^{-1}$ .

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

$k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}$

$\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}$

where $k$ is the rate constant, $E_{a}$ represents the activation energy of the chemical reaction, $R$ is the gas constant, $T$ is the temperature, and $A$ is the frequency factor.

The frequency factor, $A$ , is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be $k_{1}$ at an initial temperature $T_{1}$ and the rate constant $k_{2}$ at a final temperature $T_{2}$ , thus

$\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)$

$\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}$

Given that $E_{a} \ = \ 26.5 \ \ \text{kJ/mol}$ , $R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}$ , $T_{1} \ = \ \left(40 \ + \ 273\right) \ K$ , $T_{2} \ = \ \left(125 \ + \ 273\right) \ K$ , and $k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}$ , therefore,

$k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}$

8 0

2 years ago

Both fructose and glucose have an empirical formula of CH2O and a molecular mass of 180.15948 g/mol. Determine the molecular for

Basile [38]

Answer:

C) C₆H₁₂O₆.

Explanation:

We can determine the molecular formula by calculating the molecular mass of different choices.

Molecular mass = ∑(no. of atoms * atomic mass).

A) CH₂O:

molecular mass = atomic mass of C + 2*atomic mass of H + atomic mass of O = (12.0 g/mol) + (2 * 1.0 g/mol) + (16.0 g/mol) = 30.0 g/mol.

B) C₃H₈O₃:

molecular mass = 3(atomic mass of C) + 8(atomic mass of H) + 3(atomic mass of O) = 3(12.0 g/mol) + 8(1.0 g/mol) + 3(16.0 g/mol) = 92.0 g/mol.

C) C₆H₁₂O₆:

molecular mass = 6(atomic mass of C) + 12(atomic mass of H) + 6(atomic mass of O) = 6(12.0 g/mol) + 12(1.0 g/mol) + 6(16.0 g/mol) = 180.0 g/mol.

D) C₈H₁₆O₈:

molecular mass = 8(atomic mass of C) + 16(atomic mass of H) + 8(atomic mass of O) = 8(12.0 g/mol) + 16(1.0 g/mol) + 8(16.0 g/mol) = 240.0 g/mol.

So, the right choice is: C) C₆H₁₂O₆.

5 0

3 years ago

True or false; when something heats up, New energy is created, and when something cool down, Energy is destroyed.

Ann [662]

Answer:

This is false.

Explanation:

This statement directly opposes the first law of thermodynamics which states that energy can neither be created nor destroyed, but can transfer from one source to another.

4 0

3 years ago

The formula for chlorine dioxide

solmaris [256]

The answer is ClO2 according to google tbh

3 0

4 years ago

Read 2 more answers