The force of attraction between two masses of different substances that are in contact

How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate

zysi [14]

Answer:

12

Explanation:

the 4 by the element symbol O multiplied by the 3 on the outside of the parentheses

5 0

3 years ago

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A sample of an ideal gas at 1.00 atm and a volume of 1.87 L was placed in a weighted balloon and dropped into the ocean. As the

Ludmilka [50]

Answer:

Volume of sample after droping into the ocean=0.0234L

Explanation:

As given in the question that gas is idealso we can use ideal gas equation to solve this;

Assuming that temperature is constant;

Lets $P_1$ and $V_1$ are the initial gas parameter before dropping into the ocean

and $P_2$ and $V_2$ are the final gas parameter after dropping into the ocean

according to boyle 's law pressure is inversly proportional to the volume at constant temperature.

hence,

$P_1V_1=P_2V_2$

P1=1 atm

V1=1.87L

P2=80atm

V2=?

After putting all values we get;

V2=0.0234L

Volume of sample after droping into the ocean=0.0234L

7 0

3 years ago

Which of the following is a chemical change?

FromTheMoon [43]

Answer:

Silver tarnishing as the silver metal reacts with sulphur is a chemical change.

Explanation:

Tarnishing is the process of chemical change occurring on the surface of objects leading to corrosion or other defects on the surface. The remaining options like dilution, eroding is a physical change where the concentration of salt and rock particles will be decreased, respectively.

Similarly for soil drying also the concentration of water will be decreased leading to a physical change from wetty or dry soil.

But the last option which indicates tarnishing of silver metal on reaction with sulphur is a chemical process as the surface of silver metal will be reacting to sulphur and leads to lose of electrons which leads to corrosion of the surface or tarnishing of silver.

4 0

3 years ago

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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]

Answer: The freezing point of solution is -0.454°C

Explanation:

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:

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