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Nataly_w [17]
3 years ago
13

How many grams of carbon would be present in carbon monoxide that contains 2.6 grams of oxygen?

Chemistry
1 answer:
Andreas93 [3]3 years ago
6 0

Answer:

1.95g

Explanation:

Find mass in CO

Find C mass

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If 0.60 L of a solution contains 6.6 g of NaBr, what is its molar concentration?
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Answer:

Molarity = moles ÷ liters

to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)

Na = 22.989769

Br = 79.904

molar mass of NaBr = 102.893769

6.6g ÷ 102.893769 = 0.064143826 moles of NaBr

0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %

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3 years ago
in their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that _____.
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4 years ago
Density is anything that has mass and takes up space.<br><br> A. True<br><br> B. False
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3 years ago
Read 2 more answers
According to the chemical formula shown below, all of the following elements are found in phosphoric acid EXCEPT
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3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
3 years ago
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