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4 years ago

The reaction below demonstrates which characteristic of a base?

Chemistry

2 answers:

Alchen [17]4 years ago

7 0

Answer:

Answer is D

Explanation:

svlad2 [7]4 years ago

5 0

Answer: option D. the ability of a base to react with a soluble metal salt.

Justification:

NaOH is a strong base, which means that in water it will dissociate according to this reaction:

NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)

On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:

CuSO₄(aq) → Cu²⁺ + SO₄²⁻

Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).

That is what the given reaction represents:

CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)

↑ ↑ ↑ ↑

soluble metal salt strong base insoluble base solube salt

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3 Feas) + 4H20lg) Foog(s) + 4H₂ (g)

Nonamiya [84]

Answer:

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7 0

4 years ago

Which element is likely to have the highest thermal conductivity

givi [52]

sodium element is likely to have the highest thermal conductivity

7 0

4 years ago

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

Part A: How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

Lorico [155]

Answer:

a) 7.0 moles of NH3

b) 61.2 g of NH3

c) 4.15 g of H2

d) 8.9 ×10^19 molecules

Explanation:

Equation of the reaction;

N2(g) + 3H2(g) ⇄NH3(g)

If 3 moles of H2 yields 1 mole of NH3

21 moles of H2 will yield 21 × 1 /3 = 7.0 moles of NH3

1 mole of N2 yields 17 g of NH3

3.6 moles of N2 yields 3.6 moles × 17 g of NH3 = 61.2 g of NH3

If 6g of H2 produces 17 g of NH3

xg of H2 will produce 11.76 g of NH3

x= 6 × 11.76/17

x= 4.15 g of H2

If 6g of hydrogen yields 6.02 × 10^23 molecules of NH3

8.86 × 10^-4g of H2 yields 8.86 × 10^-4g × 6.02 × 10^23 /6 = 8.9 ×10^19 molecules

8 0

4 years ago

Carbon dioxide and methane increases in the atmosphere are most associated by scientist with

Semenov [28]

I think the answer is A, ozone depletion

5 0

3 years ago