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3 years ago

QUESTION 4

Chemistry

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

5.3 mol CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

3.2 × 10²⁴ molecules CO₂

Step 2: Convert

$3.2 \cdot 10^{24} \ mc \ CO_2(\frac{1 \ mol \ CO_2}{6.022 \cdot 10^{23} \ mc \ CO_2} )$ = 5.31385 mol CO₂

Step 3: Check

We are given 2 sig figs. Follow sig fig rules.

5.31385 mol CO₂ ≈ 5.3 mol CO₂

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The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the

gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

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2 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

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Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem

Leto [7]

Answer : The final temperature of the copper is, $87.29^oC$

Solution :

Formula used :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

Q = heat gained = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = $0.0923cal/g^oC$

$\Delta T=\text{Change in temperature}$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC$

Now put all the given values in the above formula, we get the final temperature of copper.

$299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)$

$T_{final}=87.29^oC$

Therefore, the final temperature of the copper is, $87.29^oC$

4 0

3 years ago

PLEASE HELP, I WILL GIVE BRAINLIEST AND THANKS

jasenka [17]

Answer:

23dm

Explanation:

so you have to do 10g +13dm so the you have to times it by 3 three times

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const2013 [10]

A carbonated soft drink hs a large amount of water dissolved with ample amount of carbon dioxide. In this case, the solvent is water and carbon dioxide is the solute. In 2, for the reaction CH3COOH = CH3COO + H+, CH3COOH is a Bronsted-Lowry acid because it releases a proton which is H+.

3 0

3 years ago

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