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AlexFokin [52]
3 years ago
14

What are the coefficients to balance the following equation? ba+br=babr2

Chemistry
1 answer:
scoray [572]3 years ago
8 0
1 ba+2 br——>1 babr2

u just have to make sure u have the same number of each type of atom on either side of the equation:)
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500ml of a buffer solution contains 0.050 mol nahso3 and 0.031
nydimaria [60]

Answer:

The answers are explained below

Explanation:

a)

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

b)

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

c)

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

d)

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

e)

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

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3 years ago
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kolbaska11 [484]

Answer:

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Explanation:

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3 years ago
How many atoms are in 1.2x10^5 moles of neon?​
attashe74 [19]

Answer:

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Explanation:

The number of moles contained in a substance (n) can be calculated from this expression:

nA = n × 6.02 × 10^23 atoms

Where;

nA = number of atoms of substance

n = number of moles of substance

Avagadro's number or constant = 6.02 × 10^23 atoms

Using nA = n × 6.02 × 10^23 atoms

In this question, there are 1.2 x 10^5 moles of neon (N). The number of atoms (nA) is as follows:

nA of neon = 1.2 x 10^5 × 6.02 × 10^23

nA = 7.224 × 10^ (5 + 23)

nA = 7.224 × 10^28 atoms

The number of neon atoms in 1.2 x 10^5 is 7.224 × 10^28 atoms.

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