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Nookie1986 [14]
2 years ago
5

6. M = mol solute L solution Rewrite this expression for L solution.

Chemistry
1 answer:
cricket20 [7]2 years ago
8 0

Answer:

L = mol solute / M

Explanation:

To rewrite the equation for liters (L), you need to:

M = mol solute / L                              <----- Given equation

M x L = mol solute                             <----- Multiply both sides by L

L = mol solute / M                              <----- Divide both sides by M

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If the volume of a cylinder is 84,300 cubic centimeters then how many millimeters is this equal to
nevsk [136]

Answer:

84,300,000

Explanation:

Just multiply it by 1,000. 1cm = 10mm, and a cube is the length to a sort of third degree, so you take the 10 to its third degree as well and multiply it by 1,000, instead of 10 like you would to find how many millimeters were in an amount of centimeters. If that makes sense.

8 0
3 years ago
The ____ of the sun is the interior or center part of this star
Margaret [11]
Solar core I think... i hope it helps
5 0
3 years ago
How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?
hammer [34]

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

4 0
3 years ago
A solution was prepared by mixing 0.5000 m hno2 with 0.380 m no2-. ka = 4.58 x 10-4. calculate the ph of the solution. show work
adell [148]

pH of buffer can be calculated as:

pH=pKa+log[salt]/[Acid]

As ka = 4.58 x 10-4

Concentration of [Salt] that is NO2(-1)=0.380M

Concentration of [Acid] that is HNO2=0.500M

So, pH= -log(4.58*10^-4)+log((0.380)/0.500))

=3.21

So pH of solution will be 3.21

7 0
3 years ago
Can sombody pls help me with this​
luda_lava [24]
Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)
6 0
2 years ago
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