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3 years ago

If 80 g KCl are added to 100 g of water at 70 °C, how many grams do not dissolve

Chemistry

2 answers:

sukhopar [10]3 years ago

8 0

The answer is 2
i remember doing this in middle school lol

Allushta [10]3 years ago

5 0

Answer:

Explanation:

im in highschool

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The following pieces of evidence were found at separate explosion sites. For each item, indicate whether the explosion was more

tangare [24]

Explanation:

The major difference between low and high explosives is the rate of detonation. Low explosives detonate very slowly (less than 1,000 meters per second), whereas high explosives detonate very quickly (from 1,000 to 8,500 meters per second).

High explosives among the given list are Lead azide residues, Ammonium nitrate residues, and Scraps of primacord. Whereas Nitrocellulose residues and, Potassium chlorate residues are low explosives.

7 0

3 years ago

13. What is the volume of 17.88 mol of Ar at STP?

weeeeeb [17]

Answer:

Explanation:

1 mol of ideal gas at STP occupies 22.4 (or 22.7 depending on the convention being used for STP) liters in volume. I will use 22.4 so 17.88*22.4 = 400.5 L

4 0

3 years ago

Which statement is true about a liquid but not a gas?

vovangra [49]

Answer:

I think its D

Explanation:

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

3 0

3 years ago

Read 2 more answers

An open "empty" 2 L plastic pop container, which has an actual inside volume of 2.05 L, is removed from a refrigerator at 5 °C a

Pani-rosa [81]

2.168 L of air will leave the container as it warms

<h3>Further explanation</h3>

Given

V₁=2.05 L

T₁ = 5 + 273 = 278 K

T₂ = 21 + 273 = 294 K

Required

Volume of air

Solution

Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

$\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Input the value :

V₂=(V₁.T₂)/T₁

V₂=(2.05 x 294)/278

V₂=2.168 L

3 0

3 years ago

What is the measured component of the orbital magnetic dipole moment of an electron with (a) ml = and (b) ml = ?

faust18 [17]

ANSWER:

What is the measured component of the orbital magnetic dipole moment of an electron with the values

(a) ml=3

(b ) ml= −4

a) -278 x $10^{-23}$ J/T

b) 3.71 x $10^{-23}$ J/T

STEP-BY-STEP EXPLANATION:

a) ml= 3

Цorb,z = ml Цв = - (3) * (9.27e - 24) = -278 x $10^{-23}$ J/T

b) ml= 3

Цorb,z = ml Цв = - (-4) * (9.27e - 24) = 3.71 x $10^{-23}$ J/T

3 0

3 years ago