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lys-0071 [83]
3 years ago
8

Which of the following pairs of reactants is likely to produce a pure metal when the

Chemistry
1 answer:
storchak [24]3 years ago
3 0
1. Na3PO4+Mg(NO3)2= NaNO3 + MgPO4
So this is not the answer.
2. Al+CuCl2= AlCl3+ Cu
So this is the answer.
3. Mg+O2= MgO2
So this is not an answer
4. Cl2+KI= KCl+I2
So this is also an answer

Hence B and C are answers
You might be interested in
A 50.5 L flask contains 3.25 mol of an unknown gas at 288.6 K, what is the pressure of<br> this gas?
Ivahew [28]

Answer:

1.52atm is the pressure of the gas

Explanation:

To solve this question we must use the general gas law:

PV = nRT

<em>Where P is pressure in atm = Our incognite</em>

<em>V is volume = 50.5L</em>

<em>n are moles of gas = 3.25moles</em>

<em>R is gas constat = 0.082atmL/molK</em>

<em>And T is absolute temperature = 288.6K</em>

To solve pressure:

P = nRT / V

P = 3.25mol*0.082atmL/molK*288.6K / 50.5L

P = 1.52atm is the pressure of the gas

3 0
3 years ago
Which equation represents the combined gas law?
stealth61 [152]

Answer:

P1V1/T1= P2V2/T2

Explanation:

Combined gas law involves Boyle's law and Charles law altogether with the formula of Boyle's law as P1V1=P2V2

formula for charles law as V1/T1=V2/T2

so when combined form P1V1/T1=P2V2/T2

4 0
3 years ago
Which is an example of a chemical change
jonny [76]
The answer is baking a cake.
4 0
3 years ago
Read 2 more answers
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
Here it was the picture to that
inessss [21]

Answer:

6.

a. D

b. F

Explanation:

4 0
3 years ago
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