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inysia [295]
2 years ago
5

Select all the correct answers.

Chemistry
2 answers:
hammer [34]2 years ago
4 0
It would be a mixture
Mrrafil [7]2 years ago
4 0

Answer:

liquid and mixure

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What must happen for heat to be transferred by thermal conduction?
zaharov [31]

Answer:

yh what he said

Explanation:

5 0
3 years ago
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How would you prepare a 35 ml solution of 95% (volume/volume solution of ethanol?
Artyom0805 [142]

Explanation:

<em><u>Solutions. 1. If 47 g of KCl dissolved in enough water to give 375 mL of soloution, what is the molarity ... vo volume of solute . ... v/v ethanol, how much 95% v/v ethanol ... prepare 200. mL ...</u></em>

7 0
2 years ago
At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution
IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = x_{A}\times P_{A}^{0}

vapour pressure of a component (B) in solution = x_{B}\times P_{B}^{0}

Where x_{A},x_{B} are mole fraction of component A and B in solution respectively

P_{A}^{0},P_{B}^{0} are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = 0.340\times 745 torr

                                                                         = 253 torr

7 0
3 years ago
How do you write a chemical equation given a word equation ? (Chemistry)
earnstyle [38]
Step one: Identify reactants and products and place them in a word equation.

Step two: Convert the chemical names into chemical formulas. Place them based on the chemical equation and write the state symbols.

Step three: balance the chemical equation.
5 0
3 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
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