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Xelga [282]
3 years ago
6

What is the mass in grams of 5.25 x 1024 molecules of NaOH?

Chemistry
1 answer:
Colt1911 [192]3 years ago
4 0

Answer: The mass is 348.8g

Explanation:

We begin by using Avogadro's number to convert the number of molecules of  Sodium Hydroxide to moles.

6.02 x 10∧23 molecules of NaOH -------> 1 mole of NaOH

∴ 5.25 x 10∧24 molecules of NaOH -------> 1/ 6.02 x 10∧23 x 5.25 x 10∧24 =

                                                                    8.72moles.

Having found the number of moles, we then move from moles to gram by using the molar mass of NaOH (i.e the mass of 1 mole of NaOH).

1 mole of NaOH is equivalent to 40g (Molar Mass)

8.72 moles of NaOH would be equivalent to; 40/ 1  x 8.72 = 348.8g

The mass in 5.25 x 10∧24molecules of NaOH is 348.8g

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Which redox reaction would most likely occur if silver and zinc metal were added to a solution that contained silver and zinc io
nadezda [96]

Answer:

D. Zn²⁺ + 2Ag → Zn + 2Ag⁺

Explanation:

On the reduction potential chart, we have that silver ion Ag⁺ is a stronger oxidizing agent than, zinc, Zn, which is a reducing reducing agent the compared to silver

Therefore, the redox reaction that will occur is that the zinc, Zn, will be oxidized to Zn²⁺ ion, while the silver, Ag²⁺ ion will be reduced to silver deposits Ag, therefore, the zinc will displace the silver in the solution containing silver and zinc ions because zinc is higher than silver in the reactivity series

The reduction potential Zn → Zn²⁺ + 2e⁻ = +0.76

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Therefore, given that redox reactions are reversible, we get the following likely redux reaction;

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Therefore, from the given options;

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8 0
3 years ago
Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2
Luba_88 [7]

Answer:

ΔH° =   182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

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dividing by two :

YF3(s) + 3/2H2(g)  →  Y(s) + 3HF(g)     ΔH° =  905.5 kJ/mol  Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g)  →  Y(s) + 3HF(g)     ΔH° =  905.5 kJ/mol  Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g).   ΔH° =  905.5 + (-723.1) kJ/mol

ΔH° =   182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0
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