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3 years ago

What is the ph of a 0.135 m aqueous solution of potassium acetate, kch3coo? (ka for ch3cooh = 1.8×10-5)?

1 answer:

6 0

[OH-] = √(Cs×Kw)/Ka
[OH-] = √(0,135×10^-14)/1,8×10^-5
[OH-] = √0,075×10^-9 = √75×10^-12 = 8,66×10^-6

pOH = -log[OH-]
pOH = -log[8,66×10^-6]
pOH ≈ 5,0625

pH + pOH = 14
pH = 14-5,0625
pH = 8,9375

:•)

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The chemical formula for aluminum hydroxide is Al(Oh)3. How many hydrogen atoms are present in 1 molecule here?

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There are three molecules of hydrogen.

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3 years ago

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

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Answer:

$V=0.0310L=3.10mL$

Explanation:

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In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

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$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$

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3 years ago

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WINSTONCH [101]

Answer:

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3 years ago

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Calculate the volume occupied by 272g

cupoosta [38]

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

Explanation:

Given:

mass of methane( $CH_{4}$ ) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of methane( $CH_{4}$ ) = 16.0 grams

We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

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3 years ago

Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of

saul85 [17]

Answer:

84.8%

Explanation:

Step 1: Given data

Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

Step 2: Calculate the percent yield.

We will use the following expression.

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%yield = 3.45 g / 4.07 g × 100% = 84.8%

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3 years ago