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3 years ago

The experimental percentage of oxygen in Ca(ClO3)2

Chemistry

1 answer:

Allisa [31]3 years ago

5 0

Answer:

46.379%

hope this helps :)

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Baking Soda is NaHCO3 ---Sodium hydrogen Carbonate

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What is the main cause of global conversation currents

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3 years ago

The instruction booklet for your pressure cooker indicates that its highest setting is 11.8 psipsi . You know that standard atmo

zavuch27 [327]

<u>Answer:</u> The temperature at which the food will cook is 219.14°C

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?$

Putting values in above equation, we get:

$\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$492.14=T(^oC)+273\\\\T(^oC)=219.14^oC$

Hence, the temperature at which the food will cook is 219.14°C

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3 years ago

What is the percent yield for a reaction if we predicted the formation

bogdanovich [222]

Answer:

18.1%

Explanation:

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3 years ago

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what

melomori [17]

Answer:

$Y=65.7\%$

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

$n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2$

Thus, we compute the percent yield as:

$Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%$

Best regards.

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3 years ago