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4 years ago

13

Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients NH3(g) + O2(g) + CH4(g)

→ HCN (aq) + H2O(1)

1 answer:

alexira [117]4 years ago

5 0

Answer:

$2NH_{3 (g)} + 3O_{2 (g)} +2CH_{4 (g)} --> 2HCN_{(aq)} + 6H_{2}O_{(l)}$

Explanation:

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2 years ago

How many grams of Cl are in 535 g of CaCl₂?

Sliva [168]

The answer is 341.7 g.

(1) Calculate the molar mass (M) of CaCl2 which is the sum of atomic masses (A) of elements:
M(CaCl2) = A(Ca) + 2A(Cl)
A(Ca) = 40.1 g/mol
A(Cl) = 35.45 g/mol

M(CaCl2) = 40.1 + 2 * 35.45 = 40.1 + 70.9 = 111 g/mol

(2) Calculate in how many moles are 535 g:
M(CaCl2) = 111 g/mol
111g : 1mol = 535 g : xmol
x = 535 g * 1mol : 111g = 4.82 mol

(3) Calculate how many grams of Cl are in 4.82 mol:
A(Cl) = 35.45 g/mol
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7 0

3 years ago

The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. an anal

shutvik [7]

Answer:

Amount of mercury is 1.0*10⁻⁵ g

Explanation:

<u>Given:</u>

Mercury content of stream = 0.68 ppb

volume of water = 15.0 L

Density of water = 0.998 g/L

<u>To determine:</u>

Amount of mercury in 15.0 L of water

<u>Calculation:</u>

$1 ppb = \frac{1\mu g(solute)}{1L(solvent)}$

where 1 μg (micro gram) = 10⁻⁶ g

0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water

Therefore, the amount of mercury in 15.0 L water would be:

$=\frac{0.68*10^{-6}g\ Hg* 15.0\ L\ water}{1\ L\ water} =1.02*10^{-5}g$

4 0

3 years ago