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4 years ago

Which of these can be classified as a pure substance

2 answers:

8 0

Salt can be classified as a pure substance as it is only composed of sodium chloride
while contain many other substances

schepotkina [342]4 years ago

6 0

Salt is a pure substance.
Hope it helped!

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What volume does 0.482 mol of gas occupy at a pressure of 719 mmHg at 56 ∘C?

kozerog [31]

Answer:

The volume is 13, 69 L

Explanation:

We use the formula PV=nRT. We convert the temperature in Celsius into Kelvin and the pressure in mmHg into atm.

0°C= 273K---> 56°C= 56 + 273= 329K

760 mmHg----1 atm

719 mmHg----x= (719 mmHgx 1 atm)/760 mmHg= 0,95 atm

PV=nRT ---> V= (nRT)/P

V=( 0,482 molx 0,082 l atm/K mol x 329K)/0,95 atm

7 0

4 years ago

Pickles are cucumbers preserved in a brine solution containing dill, alum, and salt. Cucumbers shrink to a fraction of their nor

Anna [14]

Answer:

hypertonic solution

Explanation:

Hypertonic solution -

It is the solution, with more amount of solute than the solvent , is known as hypertonic solution.

Now, is some substance is immersed in such solution , the substance gets shrinked , because , the solvent from the substance moves out of it and moves to the hypertonic solution.

Hence, the pickles gets shrinked up , as they put in a hypertonic solution.

8 0

3 years ago

You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.

Let's plug in the values in the equation:

$1.25M(363mL)=0.50M(V_2)$

$V_2=\frac{1.25M(363mL)}{0.50M}$

$V_2=907.5mL$

Volume of water added = 907.5mL - 363mL = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0

3 years ago

How high did Joseph Kittinger get up into the atmosphere before he jumped?

Trava [24]

I know the answers I do

8 0

3 years ago

Read 2 more answers

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis

weqwewe [10]

Answer:

$\Delta H_{rxn}=-2043.999kJ$

Explanation:

$\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]$

Where $n_{i}$ and $n_{j}$ are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

$\Delta H_{f}^{0}$ is standard heat of formation and $\Delta H_{rxn}^{0}$ is standard enthalpy change for reaction at $25^{0}\textrm{C}$

So, $\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

or, $\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]$

or, $\Delta H_{rxn}=-2043.999kJ$

4 0

4 years ago