Ask question

Ask question

All categories

Karo-lina-s [1.5K]

3 years ago

9

Finely ground nickel(II) hydroxide is placed in a beaker of water. It sinks to the bottom of the beaker and remains unchanged. A

n aqueous solution of hydrochloric acid (HCl) is then added to the beaker, and the Ni(OH ) 2 disappears. Which equation best describes what occurred in the beaker? A. Ni(OH ) 2 (s) + HCl(aq) → NiO(aq) + H 2(g) + HCl(aq) B. Ni(OH ) 2 (s) + 2HCl(aq) → NiC l 2 (aq) + 2 H 2 O(l) C. Ni(OH ) 2 (s) + 2 H 2 O(l) → NiC l 2 (aq) + 2 H 2 O(l) D. Ni(OH ) 2 (s) + 2 H 2 O(l) → NiCl 2(aq) + 3H 2O(l) + O 2(g)

1 answer:

Paraphin [41]3 years ago

5 0

Answer:C

Explanation:

You might be interested in

How do you find the atomic mass mass number of an atom?

fomenos

To calculate the atomic mass<span> of a single </span>atom<span>, add up the </span>mass<span> of protons and neutrons.

So, your answer would be A) add the number of protons and neutrons</span>

7 0

3 years ago

An orbital would never exist in the quantum description of an atom is

Mamont248 [21]

Hydrogen i think. the orbital of hydrogen is peculiar

6 0

3 years ago

What is the volume, in liters, of 4.00 moles of methane gas, CH4, at 13 ∘C and 1.20 atm?

dimulka [17.4K]

Answer:

78.268L

Signifigant figures = 78L

6 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

In cellular chemical pathways, the product(s) of any particular reaction are often quickly consumed by the next reaction in the

mezya [45]

Answer:

Tend to keep the product concetration <u>low</u> and therefore drive the reaction <u>righward</u>

Explanation:

The fact the products of a reaction are quickly consumed by the next one would tend to keep the product concetration low and therefore drive the reaction righward (to the products).

This happens because the system will not achive equilibrium between the reactants and the product, and will keep producing it util the system achives equilibrium or the reactants dry out.

5 0

3 years ago