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3 years ago

Convert eachh into scientific notation

Chemistry

2 answers:

mixer [17]3 years ago

7 0

Answer:

0.00580 →

3000 →

0.000908 →

200. →

Explanation:

yay

Ierofanga [76]3 years ago

6 0

Answer:

$\large \boxed{\mathrm{view \ explanation}}$

Explanation:

4.060 × 10⁵ (scientific notation)

The decimal point moves 5 places to the right.

⇒ 406000 (standard form)

7 × 10³ (scientific notation)

The decimal point moves 3 units to the right.

⇒ 700 (standard form)

5.0 × 10⁻⁴ (scientific notation)

The decimal point moves 4 units to the left.

⇒ 0.0005 (standard form)

8 × 10⁻² (scientific notation)

The decimal point moves 2 units to the left.

0.08 (standard form)

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

IRINA_888 [86]

The mass of a gold bar = 1.424 x 10⁴ gram

<h3>Further Explanation</h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

Known variable

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

Asked

the mass of a gold bar

Answer

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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4 years ago

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Delicious77 [7]

Answer:

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3 years ago

Calculate the mass of carbon dioxide released when 185.000 g of copper carbonate [molar mass = 123.5 g/mol] is heated (assuming

Vinvika [58]

Answer:

The answer to your question is: 65.9 g released of CO2

Explanation:

MW CO2 = 44 g

MW CuCO3 = 123.5 g

CO2 released = ?

CuCO3 = 185 g

CuCO3 ⇒ CO2 + CuO

123.5 ----------- 44g

185 g ----------- x

x = (185 x 44) / 123.5

x = 65.9 g released of CO2

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3 years ago

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pogonyaev

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