All categories

3 years ago

How many grams of salt are in 5 l of a solution with a concentration of 3 g/L

Chemistry

1 answer:

inn [45]3 years ago

8 0

Answer: 15 grams of salt are there in 5 L of solution.

Explanation:

Concentration of a solution is defined as the number of grams of solute dissolved per liter of the solution.

Given : Concentration of solution : 3 g/L

1 Liter of solution contains = 3 gram of salt

Thus 5 L of solution contains = $\frac{3g}{1L}\times 5L=15g$

Thus 15 grams of salt are there in 5 L of solution.

You might be interested in

Write the electron configuration for a sodium cation with a charge of 1+

Dafna1 [17]

$1{s}^{2} 2 {s}^{2} 2 {p}^{6}$

or simply 2,8 it is isoelectronic with argon

4 0

3 years ago

PLEASE ANSWER THIS FAST

Fofino [41]

Answer:

A and c MgSO4 and HNO3

8 0

3 years ago

Read 2 more answers

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago

Giving brainiest to whoever answers correctly.

Digiron [165]

Answer:250

Explanation:

5 0

3 years ago

Use the table on the right to calculate each required quantity.

k0ka [10]

a. 35.8 KJ

b. 871 g

c. Fe

..........................................................................................

3 0

3 years ago