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2 years ago

How many equivalents are in 0.60 mole of Mg2+

1 answer:

5 0

Equivalents express the combining power of an element for example what mass of an element can combine with or displace one gram of hydrogen or eight grams of oxygen.

Since Mg is divalent, one gram of Mg2+ can combine with 2 grams of hydrogen

Therefore, 0.6 mole of Mg2+ can combine with 0.6 x 2 = 1.2 moles of hydrogen

Hence there are 1.2 equivalents in 0.60 mole of Mg2+

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The greatest amount of energy released per gram of reactants occurs during a(1) redox reaction

LenKa [72]

The greatest amount of energy released per gram of reactants occurs during a (1) redox reaction, although it should be noted that there are exceptions depending on environment.

5 0

2 years ago

I NEED HELP ASAP, WILL MARK BRAINLEST!

Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

6 0

2 years ago

1. What is the oxidation number for the silver ion in tarnish?

spayn [35]

Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1

7 0

2 years ago

A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n

Advocard [28]

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

8 0

2 years ago

II. Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.86 g of magnesium ribbo

Leto [7]

Answer:

$Excess=3.53g$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg(s)+O_2(g) \rightarrow 2MgO(s)$

Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:

$n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO$

It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:

$m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg$

Thus, the mass in excess is:

$Excess=3.86g-0.334g\\\\Excess=3.53g$

Regards!

5 0

2 years ago