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3 years ago

How many moles of magnesium are needed to react with 16.0 g of O2?

Chemistry

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

1 mol of Mg

Explanation:

2Mg + O₂ → 2MgO

2 moles of magnesium react with 1 mol of oxygen, to produce 2 mol of magnesium oxide.

16 g / 32 g/m = 0.5 moles of O₂

If 1 mol of oxygen react with 2 mol of Mg

0.5 moles of O₂ will react with 1 mol of Mg. (0.5 ₓ2)

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Select the correct answer from each drop-down menu. consider the substances hydrogen (h2), fluorine (f2), and hydrogen fluoride

nasty-shy [4]

The boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

<h3>What is the boiling point?</h3>

The boiling point is the temperature at which the pressure exerted by the surroundings upon a liquid is equalled by the pressure exerted by the vapour of the liquid.

$F_2$ has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between $H^+$ and $F^-$ ions.

Only London Forces are formed - Therefore more energy is required to break the intermolecular forces in HF than in the other hydrogen halides and so HF has a higher boiling point.

$H_2$ and $F_2$ will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

Hence, the boiling point of HF is higher than the boiling point of $H_2$ , and it is higher than the boiling point of $F_2$ .

Learn more about the boiling point:

brainly.com/question/25777663

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7 0

2 years ago

The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

4 0

3 years ago

The mole fraction of iodine, i2, dissolved in dichloromethane, ch2cl2, is 0.115. what is the molal concentration, m, of iodine i

german

The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:

0.115 mol I₂

1 - 0.115 = 0.885 mol CH₂Cl₂

We need moles of solute, which we have, and must convert our moles of solvent to kg:

0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂

We can now calculate the molality:

m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂

The molality of the iodine solution is 1.53.

5 0

3 years ago

Atoms of the same element can have different properties.<br><br> True<br> False

Pavel [41]

I think the answer is true if so i hope it helps

5 0

3 years ago

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