Round to 4 significant figures 4,567,986

Chemistry

2 answers:

kipiarov [429]3 years ago

7 0

4,568,000
there is the rounded of 4 figures

abruzzese [7]3 years ago

4 0

4568000 ( horizontal line over 8 )

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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

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How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

HCl: 2 moles
Na: 1 mole
NaCl: 2 moles
H₂: 1 mole

You know the following masses of each element:

H: 1 g/mole
Cl: 35.45 g/mole
Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
Na: 23 g/mole
NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

HCl: 2 moles* 36.45 g/mole= 72.9 g
Na: 1 mole* 23 g/mole= 23 g
NaCl: 2 moles* 58.45 g/mole= 116.9 g
H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

$mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}$