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1 year ago

Which substance can be described as cations bonded together by mobile electrons?a. S8(s)b. Ag(s)c. SiO2(s)d. KCl(s)e. Kr(l)

Chemistry

1 answer:

Alex73 [517]1 year ago

5 0

Answer:

Ag(s)

Explanation:

Metal atoms are held together by the metallic bond. The metallic bond consists of cations held together by a sea of mobile electrons.

Thus, the crystal of Ag(s) is composed of silver ions held together by a sea of mobile electrons. Electrostatic forces of attraction holds the crystal together.

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Oxygen _____.

Marysya12 [62]

The answer is it breaks down food into energy
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1 year ago

Read 2 more answers

A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na

Leona [35]

Answer: pH=12.69

Explanation:

${\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of HF}=0.20M\times 0.6L=0.12 moles$

$HF\rightarrow H^++F^-$

Initial 0.12 0 0

Eqm 0.12-x x x

$K_a=\frac{[H^+][F^-]}{[HF]}$

$3.5\times 10^{-4}=\frac{x^2}{0.12-x}$

(neglecting small value of x in comparison to 0.12)

$x=4.2\times 10^{-5}$

Moles of $H^+=4.2\times 10^{-5}$

$NaOH\rightarrow Na^++OH^-$

${\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}$

${\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles$

0.06 moles of NaOH will give 0.06 moles of $[OH^-]$

Now $4.2\times 10^{-5}$ moles of $OH^-$ will be neutralized by $4.2\times 10^{-5}$ moles of $H^+$ and $(0.06-4.2\times 10^{-5})=0.059$ moles of $OH^-$ will be left.

Molarity of $OH^-=\frac{0.059moles}{1.2L}=0.049M$

$pOH=-\log[OH^-]=-\log[0.049]=1.31$

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0

2 years ago

A molecule contains 23.24 g iodine (I),

gavmur [86]

Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN

7 0

1 year ago

An important reaction sequence in the industrial production of nitric acid is the following: N2(g) + 3H2(g) → 2NH3(g) 4NH3(g) +

frutty [35]

Answer:

The answer to your question is 50 moles of O₂

Explanation:

Balanced Chemical reactions

1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)

2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)

moles of N₂(g) = 20 moles

moles of O₂(g) = ?

Process

1.- Calculate the moles of NH₃

1 mol of N₂ ------------- 2 moles of NH₃

20 moles of N₂ --------- x

x = (20 x 2) / 1

x = 40 moles of NH₃

2.- Calculate the moles of O₂

4 moles of NH₃ -------------- 5 O₂

40 moles of NH₃ ------------ x

x = (40 x 5) / 4

x = 200 / 4

x = 50 moles of O₂

3 0

2 years ago

When CO2(g) reacts with H2(g) to form C2H2(g) and H2O(g), 23.3 kJ of energy are absorbed for each mole of CO2(g) that reacts. Wr

kupik [55]

Answer:

$2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

Explanation:

As energy is absorbed therefore it is an endothermic reaction. Hence energy value should be written in the product side with a negative sign.

Reaction: $CO_{2}(g)+H_{2}(g)\rightarrow C_{2}H_{2}(g)+H_{2}O(g)$

C balance: $2CO_{2}(g)+H_{2}(g)\rightarrow C_{2}H_{2}(g)+H_{2}O(g)$

H and O balance: $2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)$

Here 2 moles of $CO_{2}$ react. So, energy absorbed during the reaction is $(2\times 23.3)$ kJ or 46.6 kJ

Energy balance: $2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

Balanced thermochemical equation:

$2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

3 0

1 year ago