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3 years ago

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What is the volume in liters that is needed to create a 2.3 M solution with 3.7 moles of Fe(OH)2? Round your answer to 1 decimal

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1 answer:

kolbaska11 [484]3 years ago

5 0

Answer:

Explanation:

Não sei desculpa

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If the observed value for a density is 0.80 g/mL and the accepted value is 0.70 g/mL what is the percent error?

kvv77 [185]

Answer:

<h2>The answer is 14.29 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 0.70 g/mL

error = 0.8 - 0.7 = 0.1

So we have

$P(\%) = \frac{0.1}{0.7} \times 100 \\ = 14.285714...$

We have the final answer as

<h3>14.29 %</h3>

Hope this helps you

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3 years ago

Nitrogen in plants and soil is changed into nitrogen compounds by _____.

grandymaker [24]

Answer:

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Explanation:

because It's the process by which atmospheric nitrogen is converted either by a natural or an industrial means to form of nitrogen such as ammonia.

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2 years ago

A 0.24g sample of compound of oxygen and boron was found by analysis to contain 0.144g of Oxygen.calculate the percentage compos

oksano4ka [1.4K]

Answer:

Mass of compound = 0.24 g and, mass of boron = 0.096 g percentage of boron in the compound = mass of boron / mass of compound * 100 = 0.096/0.24 * 100 = 40% mass of oxygen = 0.144 g again, mass of compound = 0.24 g percentage of oxygen in compound = mass of oxygen/mass of of compound * 100 = 0.144/0.24 * 100 = 60%

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2 years ago

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fenix001 [56]

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3 years ago

Read 2 more answers

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

2 years ago