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vladimir2022 [97]

3 years ago

9

QUICK!! WILL MARK BRAINLIEST!! what is the volume of 2.7 moles of co2 ?

1 answer:

Kay [80]3 years ago

3 0

Answer:

60. L CO₂

Explanation:

Step 1: Define

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm

Step 2: Use Dimensional Analysis (Assuming we are dealing with STP)

$2.7 \hspace{3} mol \hspace{3} CO_2(\frac{22.4 \hspace{3} L \hspace{3} CO_2}{1 \hspace{3} mol \hspace{3} CO_2} )$ = 60.48 L CO₂

Step 3: Simplify

We are given 2 sig figs.

60.48 L CO₂ ≈ 60. L CO₂

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Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

22. Most plates have a convergent boundary on one side and a

Alex73 [517]

Answer:

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8 0

3 years ago

Can someone please help!!! Its for chemistry and i looked everywhere even my text book and got no answer

Tatiana [17]

Which question the top part, middle or the 7 covalents?

8 0

3 years ago

Q8. The titration of 15.00 mL of HBr solution of unknown concentration requires 18.44 mL of a 0.100 M KOH solution to reach the

lara [203]

Answer: 0.123 M

Explanation:

According to the neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HBr$ solution = ?

$V_1$ = volume of $HBr$ solution = 15.00 ml

$M_2$ = molarity of $KOH$ solution = 0.100 M

$V_2$ = volume of $KOH$ solution = 18.44 ml

$n_1$ = valency of $HBr$ = 1

$n_2$ = valency of $KOH$ = 1

$1\times M_1\times 15.00=1\times 0.100\times 18.44$

$M_1=0.123$

Therefore, the concentration of the unknown HBr solution is 0.123 M

8 0

3 years ago

Calculate the [OH–] if the hydrogen ion concentration is 3.64 x 10–8 M.

KIM [24]

Hello.

He have that:

[H+][OH-] = 10⁻¹⁴

3.64 x 10⁻⁸ [OH-] = 10⁻¹⁴

[OH-] = 0.27 x 10⁻⁶ mol/L

3 0

3 years ago