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RoseWind [281]
2 years ago
11

How do I do number 64?

Chemistry
1 answer:
shutvik [7]2 years ago
5 0
They are isotopes because isotopes have the same number of protons (atomic number) but can have different numbers of neutrons + protons (atomic mass).
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A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

#SPJ4

7 0
1 year ago
Use the solubility generalizations on the information page to predict if one or more precipitates will form when aqueous solutio
neonofarm [45]

Answer:

3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)

Explanation:

In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>

The reaction is:

<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>
6 0
2 years ago
Calculate the mass of 4.60 x 10^25 atoms of neon?
enyata [817]
I think it’s B 5.54 x 10^2g
4 0
3 years ago
Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of microgra
leonid [27]

Answer:

0.125 mg

Explanation:

<em>The correct answer would be 0.125 mg</em>

<u>According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.</u>

milligram = 10^{-3}

microgram = 10^{-6}

Hence,

1 milligram = 1000 micrograms or 1 microgram = 10^{-3} milligram

Therefore, 125 micrograms will be:

  125/1000 = 0.125 milligram

4 0
3 years ago
How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
aksik [14]

Complete Question:

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

Answer:

2.23x10⁶ g

Explanation:

The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:

0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³

0.7976 g/m³

The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:

A = πR², where R is the radius = 1.01x10² m (half of the diameter)

A = π*(1.01x10²)²

A = 32047 m²

The volume is then:

V = 32047 * 87.32

V = 2.7983x10⁶ m³

The mass of the F⁻ is the concentration multiplied by the volume:

m = 0.7976 * 2.7983x10⁶

m = 2.23x10⁶ g

8 0
3 years ago
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