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2 years ago

324.55 cm - (6104.5 cm²/22.3 cm)

Chemistry

1 answer:

Ganezh [65]2 years ago

3 0

Answer: 50.806 cm is the correct answer.

Explanation: First divide 6104.5 cm^2 by 22.3 cm.

$\frac{6104.5cm^2}{22.3cm} = 273.74 cm$

*Note: When dividing units, subtract the exponents, and when multiplying units simply add the exponents.

Then continue by subtracting 324.55 cm - 273.74 cm.

This should give you an answer of 50.806 cm.

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SOMEONE PLEASE HELP

Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

$= 8.16 \times 10^{ - 6} \: kg$

$\boxed{density = \frac{mass}{volume} }$

Density of iron bar

$= \frac{0.0642}{8.16 \times 10^{ - 6} }$

= 7870 kg/m³ (3 s.f.)

$\boxed{mass = density \: \times volume}$

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

$= 12 \times 10^{ - 6} \: m^{3} \\ = 1.2 \times 10^{ - 5} \: m ^{3}$

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

$= \frac{0.151} {1.25 \times 10^{ - 5} } \\ = 12600 \: kg/ {m}^{3}$

(3 s.f.)

6 0

3 years ago

Calculate the % Oxygen in NaHCO3

Alexxx [7]

Answer:

Molar mass of NaHCO3 = 84.00661 g/mo

Explanation:

This compound is also known as Baking Soda or Sodium Bicarbonate.

Convert grams NaHCO3 to moles or moles NaHCO3 to grams

Molecular weight calculation:

22.98977 + 1.00794 + 12.0107 + 15.9994*3

3 0

3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago

19. What is sthe mass of 6.02 x 1024 atoms

alekssr [168]

The mass of magnesium in $6.02 \times 10^{24}$ atoms is 240 g.

Answer: Option A

<u>Explanation:</u>

First, we have to convert the atoms to moles of magnesium.

We know that $6.02 \times 10^{24}$ atoms are present in 1 mole of magnesium. So,

$\text { 1 atom }=\frac{1}{6.02 \times 10^{24}} \text { moles }$

$6.02 \times 10^{24} \text { atoms }=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}} \text { moles }=10 \text { moles }$

Thus,

$Mass of M g= molar mass of M g \times No. of moles of M g$

$\text { Mass of } M g=24 \times 10=240 g$

Thus, 240 g of Magnesium is present in $6.02 \times 10^{24}$ atoms.

4 0

3 years ago

What is the empirical formula of Fe?

Lera25 [3.4K]

Answer:

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio: The empirical formula is Fe2O3.

7 0

3 years ago

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