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3 years ago

Ni(OH)3 + Pb(SO4)2 →__Pb(OH)4 + Ni2(SO4)3 what is the balance

Chemistry

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

4+3=3+3+2

Explanation:

4Ni(oh)3+3pb(so4)2=3pb(oh)4+2Ni2(so4)3

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A(g) + 2B(g) → C(g) + D(g)

WITCHER [35]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

A(g) + 2 B(g) ⇄ C(g) + D(g)

I 1.00 1.00 0 0

C -x -2x +x +x

E 1.00-x 1.00-2x x x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

6 0

3 years ago

What name should be used for the ionic compound Cu(NO3)2

Sergeu [11.5K]

<span>Copper(II) nitrate. Hope i cleared your doubt</span>

6 0

2 years ago

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

A neutral atom posseses an atomic number of 15 and an atomic mass of 31 3 electrons are gained

kodGreya [7K]

Answer:I believe the correct answer from the choices listed above is option D. A

Explanation:neutral atom possesses an atomic number of 15 and an atomic mass of 31.Three electrons are gained. This conversion results to a negatively charged ion.

8 0

3 years ago

How many particles are in 13.5 grams of Beryllium ?

Marina86 [1]

Explanation:

I don't know this answer what Hydrogen Helium lithium

4 0

3 years ago

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