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2 years ago

Predict the missing component of each reaction.

Chemistry

2 answers:

Aleks [24]2 years ago

8 0

Answer:

2Na + Cl2 → 2NaCl

2MgO → 2Mg + O2

Snowcat [4.5K]2 years ago

7 0

Answer:

2NaCl

2Mg + O2

Explanation:

hope it helps

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The part in the middle with a coil of wire wrapped around a metal shaft is part of an electric motor. What is it?

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Answer:

rotor

Explanation:

The part of the motor being pointed to is called the <em>rotor</em>.

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3 years ago

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What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m

Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

$WO_3+3H_2\rightarrow W+3H_2O$

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

$M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g$

Theoretical yield of water : T

Moles of tungsten(VI) oxide = $\frac{51.5 g}{232 g/mol}=0.2220 mol$

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

$\frac{3}{1}\times 0.2220 mol=0.6660 mol$

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%$

The percent yield of a reaction is 48.05%.

7 0

3 years ago

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a

s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

Explanation:

Partial pressure of the $O_2$ gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

$p_{o_2}=K_H\times\chi_{O_2}$

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

$0.0013=34860 bar\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}$

Let the number of moles of $O_2$ gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L $n_w=\frac{1000 g}{18 g/mol}=55.55 mol$

$\chi_{O_2}=\frac{n}{n+n_w}$

$2.56\times 10^{-5}=\frac{n}{n+55.55}$

$n=1.43\times 10^{-7} moles$

$Molarity=\frac{\text{Moles of}O_2}{Volume}$

Molar concentration of oxygen gas in 1 L of water:

$=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L$

The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$ .

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Answer:

The answer would be D

Explanation:

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