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3 years ago

5

The equation below shows hydrogen reacting with oxygen to produce water. 2H2+O2>2H2O if 16mol of oxygen were reacted with exc

ess hydrogen gas, how many moles of water would be produced?

2 answers:

kati45 [8]3 years ago

7 0

In a chemical equation the coefficients represent the ration of the number of moles. therefore, if you have 16 moles of oxygen, you would have 32 moles of water.

Mnenie [13.5K]3 years ago

5 0

Answer : The number of moles of water produced would be, 32 moles.

Explanation : Given,

Moles of $O_2$ = 16 mole

The given chemical reaction is:

$2H_2+O_2\rightarrow 2H_2O$

From the balanced chemical reaction we conclude that,

As, 1 mole of $O_2$ gas react to give 2 moles of $H_2O$

So, 16 mole of $O_2$ gas react to give $16\times 2=32$ moles of $H_2O$

Thus, the number of moles of water produced would be, 32 moles.

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3 years ago

A sample of an ideal gas at 1.00 atm and a volume of 1.87 L was placed in a weighted balloon and dropped into the ocean. As the

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Answer:

Volume of sample after droping into the ocean=0.0234L

Explanation:

As given in the question that gas is idealso we can use ideal gas equation to solve this;

Assuming that temperature is constant;

Lets $P_1$ and $V_1$ are the initial gas parameter before dropping into the ocean

and $P_2$ and $V_2$ are the final gas parameter after dropping into the ocean

according to boyle 's law pressure is inversly proportional to the volume at constant temperature.

hence,

$P_1V_1=P_2V_2$

P1=1 atm

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3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

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Answer:

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Answer:

See Explanation

Explanation:

Given that;

N/No = (1/2)^t/t1/2

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t = time taken

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Hence;

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k = decay constant

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3 years ago