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Arturiano [62]
4 years ago
5

A solution with a pH of 6 has a ______ difference in H ion concentration than a solution with pH of 10.

Chemistry
1 answer:
sergij07 [2.7K]4 years ago
4 0

Answer:

9.999x10^-7M

Explanation:

First, let us calculate the concentration of the hydrogen ion in both cases:

For pH 6:

pH = —Log [H+]

6 = —Log [H+]

—6 = Log [H+]

Take the anti-log of —6

[H+] = 1x10^-6M

For pH 10:

pH = —Log [H+]

10 = —Log [H+]

—10 = Log [H+]

Take the anti-log of —10

[H+] = 1x10^-10M

The difference between the [H+] of pH 6 and [H+] of pH 10 is given by:

1x10^-6 — 1x10^-10 = 9.999x10^-7M

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1) 0.143 g Mg into atoms<br> 2) 0.101 kg Ti into atoms
Neporo4naja [7]

Answer:

The answer to your question is below

Explanation:

1) 0.143g of Mg into atoms

- Look for the atomic number of Magnesium in the Periodic table              

Atomic number = 24.31 g

-Use the Avogadro's number to find the number of atoms

                          24.31g ------------------- 6.023 x 10²³ atoms

                           0.143 g -----------------  x

                           x = (0.143 x 6.023 x 10²³) / 24.31

                           x = 8.613 x 10²² / 24.31

                           x = 3.54 x 10²¹ atoms

2) 0.101 kg of Ti into atoms

-Look for the atomic number of Titanium in the Periodic table

Atomic number = 47.87 g

-Use the Avogadro's number to find the number of atoms

                      47.87 g --------------------- 6.023 x 10²³

                      101 g    ----------------------  x

                       x = (101 x 6.023 x 10²³) / 47.87

                       x = 6.08x 10²⁵ / 47.87

                      x = 1.27 x 10²⁴ atoms                            

3 0
3 years ago
What is one way to make a positive impact on air quality to reduce air pollution​
geniusboy [140]

Answer:

using public transportation

Explanation:

With the use of public transport, there will be a reduction in the number of cars on the streets, this reduction will cause a reduction in the gases released by cars into the atmosphere, which will result in a reduction in air pollution.

5 0
3 years ago
At 14,000 ft elevation the air pressure drops to 0.59 atm. Assume you take a 1L sample of air at this altitude and compare it to
Ray Of Light [21]

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}

4 0
4 years ago
Half of the earth is in the light and the other half is in the shadow because
Elza [17]

Answer:

the sun doesn't shine on that half

Explanation:

4 0
2 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq) Sn2 (aq)→H2
Sav [38]

Answer:

SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O

Explanation:

At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.

for SO₄²⁻ the oxidation state of sulphur is +6 and H₂SO₃ the oxidation state of sulphur is +4

So balance equation is

(Reduction)      SO₄²⁻ + 4H⁺+ 2e⁻ → H₂SO₃ + H₂O.........................................(1)

(oxidation)        Sn²⁺ → Sn⁴⁺ + 2e⁻ .............................................................(2)

Adding equation 1 & 2

we get

SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O

6 0
4 years ago
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