All categories

2 years ago

What is the formula ofchromium(III) hydrogensulfate?

Chemistry

1 answer:

FromTheMoon [43]2 years ago

3 0

Answer:

Cr (HSO4)3

Explanation:

its molecular weight is 343.20 g/mol

its molecular formula can also be written as CrH3O12S3

molar mass of Cr (HSO4)3 can be calculated by following method;

atomic mass of Cr = 51.9961 u

atomic mass of H = 1 u

atomic mass of S = 32.065 u

atomic mass of O = 16 u

molar mass of Cr(HSO4)3 = 51.9961+ 1.00784×3 + 32.065×3 + 15.999×12

molar mass of Cr(HSO4)3 =51.9961+3.02352+96.195+ 191.988

molar mass of Cr(HSO4)3 = 343.20 g/mol

You might be interested in

Chemitey half equations

shepuryov [24]

Isn't it a because in b at the start of the equation the E in Fe just disappeared

8 0

2 years ago

What is the theoretical yield of Na2SO4, in grams?

Allushta [10]

Im sorry i dont know im sure someone will

3 0

2 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

3 0

2 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur

Brums [2.3K]

Answer:

$T_2= 31.9\°C$

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

$-Q_{rxn}=Q_{water}$

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

$-(-112000J)=m_{water}C_{water}(T_2-T_1)$

In such a way, it is possible to define the final temperature as shown below:

$T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C$

Best regards!

4 0

2 years ago

Which of the following is a physical property of nonmetals?

Minchanka [31]

The answer is (B) non metals are not very good at coducting electricity.
Hope this helps :).

5 0

3 years ago