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Kaylis [27]
3 years ago
14

Wendy made the following diagram shown to represent the solar eclipse.

Chemistry
2 answers:
DENIUS [597]3 years ago
8 0
Because the moon is not between the sun and earth or i should say earth and the sun.
Pavlova-9 [17]3 years ago
7 0
The moon should be between the sun and Earth
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What is the difference between theme and plot?
VladimirAG [237]

Answer:

should be D

Explanation:

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What binary compound would be formed from barium ions and fluoride ions?
LenaWriter [7]
Barium fluoride (BaF2) - Also known as Barium(II) fluoride - because it's a combination of two different kinds of ions (binary = two).
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Consider the following reaction at equilibrium: 2NH3 (g) N2 (g) + 3H2 (g) Le Châtelier's principle predicts that the moles of H2
Lilit [14]

Answer:

A decrease in the total volume of the reaction vessel (T constant)

Explanation:

  • Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
  • <em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
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3 years ago
Calculate the ΔHrxn for the following
Vesnalui [34]

Answer:

2 NO (g) → N2 (g) + O2 (g)

2 NOCl (g) → 2 NO (g) + Cl2 (g)

____________________________

2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

ΔH = [90.3 kJ x 2 x -1] +  [-38.6 kJ x -1 x 2] = -103.4 kJ

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6 0
2 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

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

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

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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