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exis [7]
3 years ago
15

A little boy is playing with a plastic water bottle. He blows air into the bottle to make a whistling sound. Then, he wedges a p

lastic ball into the bottle to make a rattle. But the ball is too big—it gets stuck in the mouth of the bottle! The little boy panics and runs to his big brother for help. Luckily, his older brother just studied the principle of Boyle's law. Based on what he learned, how will he get the ball out of the bottle? by pushing down on the bottle to increase the air pressure inside the bottle , by pushing down on the bottle to increase the air pressure inside the bottle by placing the bottle in the freezer to increase the air pressure inside the bottle , by placing the bottle in the freezer to increase the air pressure inside the bottle by pushing down on the bottle to decrease the temperature of the air in the bottle , by pushing down on the bottle to decrease the temperature of the air in the bottle by placing the bottle in the freezer to decrease the mass of the air inside the bottle
Chemistry
1 answer:
Harman [31]3 years ago
6 0

Answer: its by pushing down on the bottle to increase the air pressure

Explanation: because if there will be enough air, there might be a possibiity it will pop out

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What mass of CO is needed to react completely with55.0 g of Fe2O3(s)+CO(g) yield Fe(s)+CO2(g)?
Katarina [22]

Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Gather all the information in one place</em> with molar masses above the formulas and masses below them.  

M_{r}:     159.69    28.01

              Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass/g:  55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}

3.Use the molar mass of CO to calculate the mass of CO.

\text{Mass of CO} = \text{1.033 mol CO}  \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}  

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3 years ago
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3 years ago
2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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What are the rows of the periodic table of the elements also called?
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The rows are called Periods.

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