C = 0,75 mol/dm³
V = 500mL = 500cm³ = 0,5dm³

C = n/V
n = 0,75×0,5dm³
n = 0,375 moles

M NaCl: 23+35,5 = 58,5g

1 mole ---------- 58,5g
0,375 ----------- X
X = 0,375×58,5
X = 21,9375g NaCl

:)

5 0

4 years ago

Ionic compounds and metals usually form what kind of solid

natka813 [3]

Answer:

crystalline

Explanation:

Ionic compounds and metals form crystals.

4 0

3 years ago

Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O

grin007 [14]

Answer:

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Explanation:

Solubility of solute at 20°C = 32.2 g/100 grams of water

Solute soluble in 1 gram of water = $\frac{32.2}{100}g=0.322 g$

Mass of solute in soluble in 56.0 grams of water:

$0.322\times 56.0=18.032 g$

Solubility of solute at 30°C = 70.2g/100 grams of water

Solute soluble in 1 gram of water = $\frac{70.2}{100}g=0.702 g$

Mass of solute in soluble in 56.0 grams of water:

$0.702 \times 56.0=39.312 g$

If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C

Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g

Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g

Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:

39.312 g - 18.032 g = 21.28 g

21.28 grams solute can be added if the temperature is increased to 30.0°C.

8 0

3 years ago

Calculate the pH of 1.00 X 10^-6 M HCI *

murzikaleks [220]

Answer: 6

Explanation:

To find pH you have to do -log(concentration of H+)

7 0

3 years ago