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2 years ago

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An organic compound composed of carbon and hydrogen connected only by single bonds is an ________. alkane alkene alkyne aromatic

compound alcohol

1 answer:

My name is Ann [436]2 years ago

4 0

Alkane, the others have double covalent bonds

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Hii!! need help with this last answer for chem. would appreciate the help. (question 3)

avanturin [10]

Answer:

1.023 J / g°C

Explanation:

Mass, m = 37.9 g

Temperature increase (change) = 25.0°C

Heat = 969 J

Specific heat capacity , c= ?

The relationship between these quantities is given by the equation;

H = mcΔT

c = H / mΔT

c = 969 / (37.9 * 25)

c = 969 / 947.5 = 1.023 J / g°C

4 0

2 years ago

A 5.00-cm cube of magnesium has a mass of 217.501 g. What is the density of magnesiummetal?

Doss [256]

Answer:

d = 43.5 g/cm³

Explanation:

Given data:

Mass of magnesium cube = 217.501 g

Volume of magnesium cube = 5.00 cm³

Density of magnesium cube = ?

Solution:

Formula:

d = m/v

d = density

m = mass

v = volume

by putting values,

d = 217.501 g/ 5.00 cm³

d = 43.5 g/cm³

8 0

2 years ago

PLEASEE HELP ME!!!!!

Yakvenalex [24]

Answer:

Explanation:

8.61+5.779 = 14.389 = 1.4389 × 10^1

25 - 12.5 = 1.25 x 10^1

56.35 / 13.2 = 4.2689

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1 year ago

I’m to lazy can anyone help me please

katen-ka-za [31]

Answer: what do i have to do to help.. ;-; like i need instructions and why are the words coverd off black??

;-;

Explanation:

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2 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

2 years ago