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4 years ago

When working with a volumetric pipet, one adjusts the liquid level?

Chemistry

1 answer:

Lilit [14]4 years ago

7 0

One adjust the pipet by releasing the air to lower the pressure and equilizing the liquid level

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SiCl4(l) + H2O(l) → SiO2(s) + HCl(aq)

Sveta_85 [38]

I assume you’re looking for a balanced equation.
SiCl4 + 2H2O = SiO2 + 4HCl

5 0

4 years ago

PLEASE ANSWER ASAP. Which of the following represents the chemical reaction of hydrobromic acid and chlorine to produce hydrochl

Anvisha [2.4K]

Answer:

b i think

Explanation:

3 0

3 years ago

Read 2 more answers

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing

rodikova [14]

Answer:

0.225 mol = 0.23 mol to 2 significant figures

Explanation:

Calculate the moles of oxygen needed to produce 0.090 mol of water

The equation of the reaction is given as;

2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

From the equation of the reaction;

5 mol of O2 produces 2 mol of H2O

x mol of O2 produces 0.090 mol of H2O

5 = 2

x = 0.090

x = 0.090 * 5 / 2

x = 0.225 mol

8 0

3 years ago

g A mass spectrum. The peak at mass 100 has a 8% relative abundance. THe peak with mass 85 has a 40% abundance. The peak at 71 h

Sati [7]

Answer:

The peak at mass 100 with a 8% relative abundance is the molecular ion peak

Explanation:

Molecular ion peak has the highest charge to mass ratio,

Mass of 100 is same as mass to charge ratio =100

8 0

3 years ago

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