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likoan [24]
2 years ago
12

Convert 15.2 moles of K to atoms of K.

Chemistry
1 answer:
choli [55]2 years ago
3 0

Answer:

\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

Multiply by the given number of moles: 15.2

15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

The moles of potassium cancel.

15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }

The denominator of 1 can be ignored.

15.2 * {6.022*10^{23} \ atoms \ K }{

Multiply.

9.15344*10^{24} \ atoms \ K

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

9.15*10^{24} \ atoms \ K

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

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Classify the following aqueous solutions as: strong acid, weak acid, neutral, weak base, or strong base.
kirill [66]

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4 0
1 year ago
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13
Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\

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Calculating the standardn:

\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\

Please find the correct equation in the attached file.

=28.195

For point a:

=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\

For point b:

=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M

For point c:

= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\  M

It is calculated by using the slope value that is 1.75 \times 10^9 M^{-1}. The slope value 1.75 \times 10^9 M^{-1}is ambiguous.

7 0
2 years ago
A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc
Elan Coil [88]

Answer:

a. Moles of NaOH  = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Explanation:

a.

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For NaOH :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of NaOH :

Moles=0.1052 \times {15.62\times 10^{-3}}\ moles

Moles of NaOH  = 0.001643 moles

b.

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

c.

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains \frac{0.296}{1.159}\times 1.213 g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.

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Answer:

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The difference is, however, since barium is lower in group than magnesium, it has more electron shells than magnesium and, therefore, its radius is greater. The attraction force between the nucleus and the valence electrons in barium is lower as a result, as force is inversely proportional to distance. Hence, a lower amount of energy is required to remove the valence electrons from barium.

3 0
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