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2 years ago

I WILL GIVE BRAINLIEST AND EXTRA POINTSSS! PLEASE ANSWER

Chemistry

2 answers:

Bas_tet [7]2 years ago

7 0

The dependant variable is what you measure during the experiment and what is affected in the experiment (this is for question 4) so the answer would be D

Alex2 years ago

7 0

The answer of Question 3 is d

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Atoms receive a positive charge when they take electrons. Atoms give of negative electrons.

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2 years ago

Acid rain is a type of pollution caused mainly from

Aleksandr-060686 [28]

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2 years ago

Suppose that a certain fortunate person has a net worth of $79.0 billion ($7.90×1010). If her stock has a good year and gains $3

mario62 [17]

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3 years ago

18 ) Liquid A and Liquid B are clear liquids. They are placed in open containers and allowed to evaporate. When evaporation is c

arsen [322]

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Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture

Explanation:

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2 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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2 years ago