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4 years ago

Please help me in my TEST

Chemistry

1 answer:

Natali [406]4 years ago

4 0

Answer:

D) Please look below at the cart to cheak your answer!

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The molar mass of aniline is 93 g/mol. what is its molecular formula

Mashutka [201]

Answer:

Molecular formula of aniline is C6H5NH2.

Explanation:

As we know, molecular mass can be calculated as

Molar mass = C6H5NH2

Molar mass = (6*12)+(1*7)+(28)

Molar mass = 93 g/mol

4 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

4 years ago

Identify each of the following as either an element or Compound. Write (E) for Element and (C)

castortr0y [4]

Carbon = C

Water = C

Aluminum foil = E

Plastic = E

Tin = E

Silicon dioxide = C

Helium = C

Arsenic = C

Carbon dioxide = C

Sodium Chloride = C

4 0

3 years ago

1. Methane (CH4) combusts as shown in the equation below. If 3 moles of methane were used in the combustion, what would be the c

umka2103 [35]

Answer:

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

Explanation:

Step 1: The balanced equation

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH = -802 kJ

Step 2: Given data

We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.

The enthalpy change of combustion, given here as Δ H , tells us how much heat is either absorbed or released by the combustion of one mole of a substance.

In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.

Step 3: calculate the enthalpy change for 3 moles

The -802 kj is the enthalpy change for 1 mole

The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

5 0

3 years ago

One half of a balanced chemical equation is shown. 3Mg(OH)2 + 2H3PO4 Which lists the numbers of each atom the other half of the

Fudgin [204]

Answer:

A) 3 Mg, 2 P, 14 O, 12 H

Explanation:

On edgenuity 2020

7 0

3 years ago

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