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Gekata [30.6K]
2 years ago
6

11. Which is the solubility product expression for Na3PO4(s)?

Chemistry
1 answer:
dlinn [17]2 years ago
8 0

Answer: B

Explanation: the 3 in Na3PO4(s) belong to the sodium atom (Na). so in any of these equations, the 3 would have to be with Na.

A- the 3 is along w the PO4, which would make it part of that bond

C- there is no 3 at all for Na in this choice, making it incorrect

D- Again, the 3 is placed on the other half of the bond

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Does the density of ice affect the melting rate of ice, or does adding objects affect melting rates? I'm going to need evidence
kotykmax [81]
The density of ice does not affect the melting rate. But, adding an object does affect the melt rate. The reason this is is because when there is an object, there is less to melt. Hence, affecting the melting rate.
7 0
3 years ago
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What is the percent by mass of oxygen fe2o3 (formula mass= 160)?
GarryVolchara [31]
30% should be the percentage of oxygen if the total mass of fe2o3 is 160.
5 0
2 years ago
A student was asked to determine the densities of the four metals listed. The student was asked to determine the density of an u
Iteru [2.4K]

Answer:

D is correct

Explanation:

because

we know that

density of lead is 11.36 g/cm3

and

density of tin is 7.31 g/cm3

so..

density of alloy by mixing 50/50

=(11.36+7.31)/2 g/cm3

=18.67/2 g/cm3

=9.33 g/cm3

5 0
3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
Disadvantages of Mendeleev's periodic table​
mestny [16]

Answer:

Various limitations of Mendeleev's periodic table are:-

Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .

Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.

In the d-block , elements with lower atomic number were placed before higher atomic number.

Explanation:

3 0
2 years ago
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