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4 years ago

Convert 2.8 pounds to kg

Chemistry

2 answers:

Nikitich [7]4 years ago

6 0

2.8 pounds =
1.27 kilograms

Pani-rosa [81]4 years ago

5 0

Answer:

1.27006 kg

Explanation:

1 lb is equal to 0.907185 kg

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3 years ago

Which of the following is a molecular formula for a compound with an empirical formula of CH2O and a molar mass of 150. g/mol.

trapecia [35]

Answer:

C₅H₁₀O₅

Explanation:

Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that

n = molar mass of the molecular formula / molar mass of the empirical formula

The molar mass of the molecular formula is 150 g/mol.

The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol

n = (150 g/mol) / (30 g/mol) = 5

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CH₂O × 5 = C₅H₁₀O₅

7 0

3 years ago

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

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

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

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

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

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago