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gulaghasi [49]
3 years ago
14

A compound such as CaCl, would have an overall charge of?

Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0

Calcium Chloride would have a charge of +2

Hopes this helps :)

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Compare the patterns of earthquakes, volcanoes, and mountain ranges on earth?
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Its B i already got it
Rina8888 [55]

Explanation:

With all the scaling up and design improvements, the flight range of the R-7 increased to an intercontinental distance from 330 kilometers for the V-2. The R-7 ended up looking oversized even next to its American contemporaries -- the Atlas and Titan ballistic missiles.

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what is the region surrounding the nucleus which is occupied by electrons?????? its a crossword ( _ _ _ _ _ _ _ _ C _ _ _ _)
ASHA 777 [7]

Answer:

Electron cloud

Explanation:

The electron cloud is an informal term in chemistry and physics. This region describes the negatively charged electrons in their orbit surrounding the nucleus in an atom. The nucleus has protons in it that have a positive charge on it. It is not easy to know where exactly an electron in any specific time but the electron cloud helps to know electron is moving in this specific area.

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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
3 years ago
How many licks does it take to get to the end of a tootsie pop
worty [1.4K]

3 lol i was thinking about that from the ad

4 0
3 years ago
Read 2 more answers
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