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4 years ago

What kind of thing can be measured in derived units?

Chemistry

1 answer:

sergejj [24]4 years ago

7 0

Answer:

In this SI units system, there are seven SI base units and three supplementary units. The base SI units are metre, kilogram, second, kelvin, ampere, candela and the mole and the three supplementary SI units are radian, steradian and becquerel. All other SI units can be derived from these base units.

Explanation:

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The solid fuel in the booster stage of the space shuttle is a mix- ture of ammonium perchlorate and aluminum powder. Upon igni-

nataly862011 [7]

Answer:

For A: The mass of aluminium required will be 183 g

For B: The mass of alumina produced will be $6.63\times 10^6g$

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For a:

Given mass of $NH_4ClO_4$ = 1.325 kg = 1325 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $NH_4ClO_4$ = 117.50 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

6 moles of $NH_4ClO_4$ reacts with 10 moles of aluminium

So, 11.28 moles of $NH_4ClO_4$ will react with = $\frac{10}{6}\times 11.28=6.77mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

$6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g$

Hence, the mass of aluminium required will be 183 g

For b:

Given mass of aluminium = $3.500\times 10^3=3.5\times 10^6g$ (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, $1.3\times 10^5mol$ of aluminium will react with = $\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol$ of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = $6.5\times 10^4mol$

Putting values in equation 1, we get:

$6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g$

Hence, the mass of alumina produced will be $6.63\times 10^6g$

6 0

3 years ago

Which of the following is true regarding atomic radius?

Ymorist [56]

attachment/photo please

5 0

3 years ago

A mixture of carbon dioxide and hydrogen gases is maintained in a 6.68 L flask at a pressure of 2.14 atm and a temperature of 19

matrenka [14]

Answer:

The mass of hydrogen gas in the mixture: w₂ = 0.433 g

Explanation:

According to the ideal gas equation:

for an ideal gas, $P.V = n_{total}.R.T$

and $n_{total}= n_{1}+n_{2}$

Here, P: total pressure of the gases = 2.14 atm

V: total volume of the gases = 6.68 L

T: temperature = 19 °C = 19+273.15 = 292.15K (∵ 0°C = 273.15K)

R: gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹

$n_{total}$ : total number of moles of gases

To calculate the total number of moles of gases:

$n_{total} = \frac{P.V}{R.T} = \frac{2.14 atm\times 6.68 L}{0.08206 LatmK^{-}mol^{-}\times 292.15K}$ = 0.5963 moles

Let, the number of moles of carbon dioxide be n₁ and number of moles of hydrogen be n₂

Given: mass of carbon dioxide: w₁ = 16.8 g, mass of hydrogen: w₂ = ?g

molar mass of carbon dioxide: m₁ = 44.01 g/mol, molar mass of hydrogen: m₂= 2.016 g/mol

Therefore, $n_{total}$ = n_{1}+n_{2} = (w₁ ÷ m₁) + (w₂ ÷ m₂)

⇒ 0.5963 mol = (16.8 g ÷ 44.01 g/mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol = (0.3817mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol - 0.3817mol = (w₂ ÷ 2.016 g/mol)

⇒ 0.2146 mol = (w₂ ÷ 2.016 g/mol)

⇒ w₂ = 0.433 g

Therefore, the mass of hydrogen gas in the mixture: w₂ = 0.433 g

4 0

4 years ago

A sample 0. 100 moles of a gas is collected at at stp. What is the volume of the gas in liters?

andrew11 [14]

A sample 0. 100 moles of a gas is collected at at STP . 2.24 is the volume of the gas in liters.

The STP means standard temperature and pressure.

At STP,

Temperature = 0 °C =273 K

Pressure = 1 atm

We get value of volume by using ideal gas equation,

PV = nRT

P is the pressure of the gas = 1 atm
V is the volume occupied by the gas = ?
n is the number of the moles = 1 mole
T is the temperature of the gas = 273 K or 0 °C
R universal gas constant = 8.31 J/ mole × K

Calculation,

Since, one mole of a gas occupy 22.4 L volume at STP

So, for 0.1 mole volume occupy = 22.4L × 0.1 mole/1 mole = 2.24L

To learn more about volume at STP,

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6 0

1 year ago

The stage of the cell cycle that follows mitosis is called _____.

Anon25 [30]

The answer is metaphase

3 0

3 years ago

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