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Degger [83]
3 years ago
8

3. A student measured 15.0 grams of ice in a beaker. The beaker was then

Chemistry
1 answer:
gregori [183]3 years ago
3 0

Answer:

The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules

Explanation:

The mass of ice in the beaker = 15.0 grams

The initial temperature of the ice = 0°C

The final temperature of the ice = 0°C

The latent heat of fusion of ice = 330 J/g

The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice

Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J

The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.

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Definition of proton and example
gtnhenbr [62]

Answer:

An elementary particle that is identical with the nucleus of the hydrogen atom, that along with the neutron is a constituent of all other atomic nuclei, that carries a positive charge numerically equal to the charge of an electron.

Example:

The nucleus of a hydrogen atom or the H+ ion is an example of a proton. Regardless of the isotope, each atom of hydrogen has 1 proton; each helium atom contains 2 protons; each lithium atom contains 3 protons and so on.

3 0
3 years ago
The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample
IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

             Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

                Q = 0.031 calories/gram °C x 10.0 g x 20 °C

                Q = 6.2 calories

So option A is correct = 6.2 calories

6 0
3 years ago
How many grams of calcium chloride are dissolved in 5.65 liters of a 0.11 m solution of calcium choride?
Julli [10]
C = 0.11 mol
V = 5.65 L
n = ???

n = C*V
n = 0.11 * 5.65
n = 0.622 mols

1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x

x = 111 * 0.622
x = 69.0 grams CaCl2
4 0
3 years ago
Teo3 lewis structure
svetoff [14.1K]

Answer:

here. hope it helps!

Explanation:

5 0
1 year ago
What is a bond between a positive and a negative ion called?
Marizza181 [45]
I think they are called ionic bonds.
3 0
3 years ago
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