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2 years ago

An ion is formed when an atom___?

Chemistry

1 answer:

KATRIN_1 [288]2 years ago

5 0

Answer:

gain or loses electrons.

Explanation:

Because the atom need to be balanced.

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What is the definition of relief in science terms

SOVA2 [1]

The apparent topography exhibited by minerals in thin section as a consequence of refractive index.

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3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

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2 years ago

I was named after the man that developed the periodic table.

valina [46]

Einsteinium
I hope that helps

4 0

2 years ago

Read 2 more answers

What are the properties of elements related to?

nekit [7.7K]

The correct answer is A) the number of electrons that fill the outer shell.

Brady

5 0

2 years ago

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A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

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3 years ago

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