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lana [24]
3 years ago
8

What mass of copper is produced when zinc is added to a solution containing 31.9g copper ii tetraoxosulphate vi

Chemistry
1 answer:
solmaris [256]3 years ago
4 0

Answer:

12.7g of Cu

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

Zn + CuSO4 —> ZnSO4 + Cu

Molar Mass of Cu = 63.5g/mol

Molar Mass of CuSO4 = 63.5 + 32 + (16x4) = 63.5 + 32 + 64 = 159.5g/mol

From the equation,

159.5g of CuSO4 produced 63.5g of Cu.

Therefore, 31.9g of CuSO4 will produce = (31.9 x 63.5) / 159.5 = 12.7g of Cu

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Joe tries to neutralize 125mL of 2.0 M HCL with 175mL of 1.0M NaOH. will the neutralization reaction be completed? If not what i
Lemur [1.5K]

Answer:

1) The neutralization reaction will mot be completed.

2) pH = 0.6.

Explanation:

<em>1) Will the neutralization reaction be completed?</em>

For the neutralization reaction be completed; The no. of millimoles of the acid must be equal the no. of millimoles of the base.

The no. of millimoles of 125 mL of 2.0 M HCl = MV = (2.0 M)(125.0 mL) = 250.0 mmol.

The no. of millmoles of 175 mL of 1.0 M NaOH = MV = (1.0 M)(175.0 mL) = 175.0 mmol.

∴ HCl will be in excess.

∴ The neutralization reaction will mot be completed.

<em>2) If not what is the pH of the final solution?</em>

[H⁺] =[ (MV)HCl - (MV)NaOH]/V total = (250.0 mmol - 175.0 mmol) / (300.0 mL) = 0.25 M.

∵ pH = - log[H⁺]

<em>∴  pH =</em> - log(0.25) =<em> 0.6.</em>

4 0
3 years ago
Troposphere vs stratosphere, differences
Alex Ar [27]

Answer:

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region. The stratosphere starts just above the troposphere and extends to 50 kilometers (31 miles) high.

8 0
3 years ago
Read 2 more answers
Phosphorus pentachloride decomposes at higher temperatures. PCl5(g) ⇄ PCl3(g) + Cl2(g) An equilibrium mixture at some temperatur
aliya0001 [1]

Answer:

The reaction shifts from right to left. The new concentration of PCl₅ in equilibrium is 1.133 M.

Explanation:

The stress applied to the system is the addition of Cl₂. To offset this  stress, <u>some Cl₂ reacts with PCl₃ to produce PCl₅ until a new equilibrium is established</u>.

The net reaction therefore shifts from right to left, that is,

PCl₅ (g) ← PCl₃ (g) + Cl₂ (g)

To work with the Kc expression, we need to convert the grams to moles. Since the volume of the flask is 1 liter, the moles can be expressed as molarity.

For PCl₅:

208.23 g ---------- 1 mol

3.74 g -------------- x= 1.80 x10⁻² mol ⇒ 1.80 x10⁻² M

For PCl₃:

137.33 g ---------- 1 mol

4.86 g ------------- x= 3.54 x10⁻² mol ⇒ 3.54 x10⁻² M

For Cl₂:

70.91 g ---------- 1 mol

3.59 g ----------- x= 5.06 x10⁻² mol ⇒ 5.06 x10⁻² M

Now, taking into account the shift of the equation, we make an ICE chart:

                   PCl₅ (g)       ←            PCl₃ (g) +                           Cl₂ (g)

i)             1.80 x10⁻²                   3.54 x10⁻²                           5.06 x10⁻²

c)                   +x                                 -x                             +1.31 x10⁻² -x

e)           1.80 x10⁻² +x           3.54 x10⁻² -x             5.06 x10⁻²+1.31 x10⁻² -x

Now, we write the Kc formula:

Kc = [PCl₃] [Cl₂] / [PCl₅]

Kc = [ 3.54 x10⁻² -x] [5.06 x10⁻²+1.31 x10⁻² -x] / [1.80 x10⁻² +x]

After working with the expression, we find that x = 1.115.

Therefore, the new concentration of  PCl₅ will be 1.133 M.

7 0
3 years ago
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An experiment with high validity is
uysha [10]

Answer:

extraneous variables.

Explanation:

5 0
3 years ago
Complete and balance the following equation:
sertanlavr [38]
You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.

You get: BrO3 + N2H4 -> Br2 + N2 + H2O

# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2

Br:
Multiply the reactant (left) side by 2 to balance.

O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.

H:
The product side is now equal to 12. Multiply the reactant side by 3 to balance.

N:
Now you have to balance N because the reactant side has been risen. So multiply the product side by three as well.

You end up with the complete and balanced equation:

2BrO3 + 3N2H4 -> Br2 + 3N2 + 6H2O
4 0
4 years ago
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