All categories

2 years ago

How many kg of solvent would contain 130 grams of ZnBr2 if the solution was 0.950 grams

Chemistry

1 answer:

Semenov [28]2 years ago

5 0

Answer:

0.00731 kg

Explanation:

Given that solution is a product of solute and solvent.

Hence, solution = solute * solvent

In this case, the solution is given as 0.950 grams = 0.00095 kilograms

solute is given as 130 grams = 0.13 kilograms

Hence to get the amount of solvent in kilograms, we have

Solution ÷ Solute => 0.00095 ÷ 0.13

= 0.0073076923

=> 0.00731 kilograms

You might be interested in

9.69 × 10^7 in decimal form

iren [92.7K]

The answer is 96,900,000

4 0

1 year ago

Is anybody good in Chemistry that can help me with this?

sesenic [268]

Answer:

just replace the 9 mole with 3.68 g of Al .

I think it will help you.

5 0

2 years ago

How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?

Bad White [126]

The balanced chemical reaction is :

$O_2 + 4Na \ -> \ 2Na_2O$

Number of moles of Na, $n = \dfrac{14.6}{23} = 0.635 \ mol$ .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

$n = \dfrac{0.635}{4}\ mole$

So, amount of oxygen required is :

$m = \dfrac{0.635 \times 32}{4}\ gm\\\\m = 5.08 \ gm$

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0

2 years ago

g modenr vacuum pumps make it easy to attain pressures of the order of in the laboratory. at a preasusure of 6.75 atm and an ord

Ratling [72]

Answer:

Number of molecules = 1.8267×10^20

Explanation:

From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;

PV = nRT

where

P =pressure

V =volume

n = the number of moles

R is the gas constant equal to 0.0821 L·atm/mol·K

T is the absolute temperature

Given:

P = 6.75 atm;

T = 290.0 k,

; V = 1.07 cm³ = 0.001 L

( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)

n = 3.0335167*10^-4 moles

But there are 6.022×10²³ molecules in 1 mole,

Number of molecules = 1.8267×10^20

7 0

2 years ago

Calcuate the number of<br> grams of solute in 453.9mL<br> of 0.237 M calcium acetate

AysviL [449]

The number of grams : 17.082 g

<h3>Further explanation</h3>

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large {\boxed {\bold {M ~ = ~ \dfrac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

453.9 mL of 0.237 M calcium acetate

$\tt mol=M\times V=0.237\times 0.4539=0.108$

mass

MW Ca(C₂H₃OO)₂ : 158,17 g/mol

$\tt mass=mol\times MW\\\\mass=0.108\times 158.17=17.082~g$

4 0

2 years ago