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slava [35]
3 years ago
14

Which of the following is true of liquids and solids?

Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0
The answer is D. This is because liquids take up the shape of the container they are in, so it is never definite. Where as solids stay the same shape.
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To make 2.00 L of 0.300 M sulfuric acid, how many mL of a 1.00 M stock solution should be used?
Tamiku [17]
We are given with
M1 = 1.00 M

M2 = 0.300 M
V2 = 2.00 L

We are asked to get V1

Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL<span />
3 0
3 years ago
"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"
Artyom0805 [142]

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

4 0
2 years ago
The Ksp can be used to find the concentration of hydroxide ions, and thus determine the pH given a concentration of of aluminum.
Anton [14]

Answer:

3.9.

Explanation:

(Molar mass of 342.15 g/mol).

The first thing to do in this question is to coverts our units. We are given 6.70 lb = 453.592 × 6.70 lb = 3039.0664 g of aluminum sulfate.

2050 gallons of water = 2050 × 3.785 = 7759.25 Litres.

After this we will now look for the Concentration aluminum sulfate using the formula below:

Concentration = number of moles ÷ volume. ------------------------------------(1).

Recall; Number of moles = 3039.0664/ 342.15 = 8.8823 moles. The volume = 2050 gallons = 7759.25 Litres.

Therefore, slotting in the values of of number of moles and volume into the equation (1) above, we have;

Concentration= 8.8823 moles/ 7759.25 Litres. = 0.001145 M.

The next equation we are going to make use of today is that of the solubilty product that is;

Ksp = (Al^3+) (OH^-)^3.

So, we have the ksp = 1.3 × 10^-33 and the value of (Al^3+)= 0.001145 M.

Hence, making (OH^-) the subject of the formula;

(OH^-) = ( 1.3 × 10^-33 / 0.001145)^1/3.

(OH^-) = 8 × 10^-11 M.

Hence, pOH = - log (OH^-)

pOH = - log (8 × 10^-11).

pOH = 10.1

Therefore, the values of pH is; 1

pH + pOH = 14.

pH = 14 - 10.1 = 3.9.

3 0
3 years ago
What does a horizontal row of elements in the periodic table represents
xxTIMURxx [149]
The periods or energy levels
8 0
3 years ago
Read 2 more answers
You have a 5-liter container with 1.30 x 1024 molecules of ammonia gas (NH3) at STP.
bogdanovich [222]

Answer:

3). 1.30 × 10^(24) molecules

Explanation:

From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.

We can relate it to this question as;

V₁/n₁ = V₂/n₂

Where;

V₁ is initial volume

n₁ is initial number of molecules

V₂ is final volume

n₂ is final number of molecules

Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;

5/(1.30 x 10^(24)) = 5/n₂

n₂ = 1.30 x 10^(24) molecules

6 0
2 years ago
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