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elena55 [62]
3 years ago
7

Calculate the mass in grams for a single molecule of carbon monoxide​

Chemistry
1 answer:
Oxana [17]3 years ago
4 0

Answer:

7.307 x 10^-23

Explanation:

mass of one molecule = mr/ avogadros number

= 44/6.022 x 10^23

= 7.307x 10^-23

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NaClO3 &gt; NaCl + O2 <br> Balance
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Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

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Which tool was most likely used in a procedure if the lab report shows that approximately 300 mL of water was used?
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15. What volume of CCI, (d = 1.6 g/cc) contain
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Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

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4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

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Hydrofluoric acid and Water react to form fluoride anion and hydronium cation, like this HF(aq) + H_2O(l) rightarrow F(aq) + H_3
maksim [4K]

Answer:

Kc = 1.09x10⁻⁴

Explanation:

<em>HF = 1.62g</em>

<em>H₂O = 516g</em>

<em>F⁻ = 0.163g</em>

<em>H₃O⁺ = 0.110g</em>

<em />

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>

<em />

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

<h3>Kc = 1.09x10⁻⁴</h3>
7 0
3 years ago
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