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3 years ago

Explain the difference between cocl2 6h2o and anhydrous cobalt chloride

1 answer:

8 0

Colbat (ii) which is a compound birth out of the combination of chlorine and colbat to form Cocl2.6h2o has water in it as we can see from it's chemical it's hexahydrate Anhydrous cobalt chloride as the word anhydrous clearly states , does not have water in

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Can you help plzzz thank you

iogann1982 [59]

Answer:

It would be Container 3

Explanation:

Each of the containers has the same amount of salt. Salinity refers to salt level. Since the question is asking for the container with the greatest salinity, you are looking for the container with the least water (because it will be the saltiest out of all of them). Container 3 has the least water.

Hope this helps :)

5 0

3 years ago

Read 2 more answers

PLEASE HELP!!! 890 g of reactant A completely reacts with 120 g of reactant B to form two products. If product A has a mass of 9

ohaa [14]

it's (890+120)-92 which is 918g

7 0

3 years ago

Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitro

vitfil [10]

Answer : The partial pressure of $O_2$ is, 222.93 torr

Explanation :

Half-life = 2.81 hr = 168.6 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{168.6min}$

$k=4.11\times 10^{-3}min^{-1}$

Now we have to calculate the partial pressure of $O_2$

The balanced chemical reaction is:

$2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)$

Initial pressure 760 0 0

At eqm. (760-2x) 4x x

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{P_o}{P_t}$

where,

k = rate constant

t = time passed by the sample = 215 min

a = initial pressure of $N_2O_5$ = 760 torr

a - x = pressure of $N_2O_5$ at equilibrium = (760-2x) torr

Now put all the given values in above equation, we get:

$215=\frac{2.303}{4.11\times 10^{-3}}\log\frac{760}{760-2x}$

$x=222.93torr$

The partial pressure of $O_2$ = x = 222.93 torr

7 0

3 years ago

If 1.00 calorie = 4.18kj, how many kj of energy can be released by an apple containing 125 cal?

Georgia [21]

Answer:

the answeerrrr issues c

6 0

3 years ago

Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

To calculate the molality of solution, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0

3 years ago