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3 years ago

WILL AWARD BRAINLIEST! How many moles of iron (III) oxide are produced when 0.275 moles of iron is reacted?

Chemistry

2 answers:

aev [14]3 years ago

8 0

Answer:

Arrow should be yeilds, THE COEFFICIENTS are mole ratios. So every 4 moles of NH3 or ammonia produce 6 moles of H2O water

STALIN [3.7K]3 years ago

3 0

Answer:

Solution given:

3Fe+ O2. =Fe3O2

3mole. 1 mole. 1 mole

we have

3 mole of iron gives 1 moles of 1 mole of fe3O2

now

0.275 mole of iron gives 1×0.275/3=0.92mole of fe3O2

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Which layer contains almost 90% of the atmosphere's total mass?

Afina-wow [57]

The answer is the troposphere

4 0

3 years ago

Write an equation for the hydrogenation of glyceryl trilinolenate, a fat formed from glycerol and three linolenic acid molecules

Tomtit [17]

Balanced chemical equation for the hydrogenation of glyceryl trilinolenate:

C₅₇H₉₂O₆ + 9H₂ → C₅₇H₁₁₀O₆.

Linolenic acid (octadecatrienoic acids ) is a type of fatty acid. It has 18 carbon atoms chain and three double bonds. So trilinolenate has nine double bonds.

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Hydrogenation is addition of hydrogen atoms at both sides of a double bond.

5 0

4 years ago

70 points!!!! will mark the brainliest,<br> ANSWER ASAP PLS??????????????

chubhunter [2.5K]

Answer:

I believe it's A.

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the percentage of calcium in calcium chlorate

givi [52]

Answer:

19.4%

Explanation:

Calcium Chlorate is Ca(ClO3)2

Now calculate the molar mass of Ca(ClO3)2

Ca = 40.1

Cl = 35.5

O = 16.0

But they are two chlorine atoms and six oxygen atoms. So you do this:

40.1 + 35.5(2) + 16.0(6) = 207.1 grams

Now find the molar mass of just calcium.

There is only one calcium atom.

So you do this.

40.1(1) = 40.1

Now divide the molar mass of calcium by the molar mass of calcium chlorate.

40.1 / 207.1 = 0.1936

0.1936 rounds to 0.194

Now multiply 0.194 * 100 and you will get 19.4

So the final answer is 19.4%.

Hope it helped!

4 0

3 years ago

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

$n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

$n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$

= 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

= $\frac{0.0845}{0.4469 M}$

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

= 0.283 M

Chemical equation for this reaction is as follows.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

= $\frac{10^{-14}}{10^{-3.36}}$

= $2.29 \times 10^{-11}$

Now, $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

= $\sqrt{2.29 \times 10^{-11} \times 0.283}$

= $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

pH = $-log [H_{3}O^{+}]$

= $-log (2.546 \times 10^{-6})$

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0

3 years ago