What is the reaction type<br> 2 Mgl2+ Mn(SO3)2 → 2 MgSO3 + Mnl4

A. Energy is transferred to different forms

Luden [163]

Answer:

A. Energy is transferred to different forms .

Explanation:

Hello!

In this case, we need to consider the law of conservation of mass and energy which states that mass and energy cannot be neither created nor destroyed, just modified; it means we can rule out B. and C. so far.

Moreover, since D. is actually true for combustion reactions because they are used to provide energy in industrial operations, this is not the concern here because a combustion reaction is not considered.

Therefore the correct option is A. Energy is transferred to different forms as the energy provided by Rose is transferred to the pendulum system .

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3 0

3 years ago

Which metal atom below would Not be involved in formation of a Type II Compound?

BartSMP [9]

In formation of a Type II Binary Compound, the metal atom present is<span> NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.

4 0

3 years ago

Many esters have characteristic odors. Methyl salicylate exhibits a pleasant wintergreen odor. This ester can be prepared from s

hodyreva [135]

Answer:

Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.

Explanation:

5 0

2 years ago

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)

lesya692 [45]

Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.

Explanation:

Given: Current = 62.0 A

Time = 23.0 sec

Formula used to calculate charge is as follows.

$Q = I \times t$

where,

Q = charge

I = current

t = time

Substitute the values into above formula as follows.

$Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C$

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

$Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol$

The oxidation state of Pb in $PbSO_{4}$ is 2. So, moles deposited by Pb is as follows.

$Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol$

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g$

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.

3 0

2 years ago

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be

mylen [45]

yield = 52.23 %

Explanation:

We have the following chemical reaction:

2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)

mass of aluminium = mass of bottle with aluminium pieces - bottle mass

mass of aluminium = 10.8955 - 9.8981 = 0.9974 g

mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if 2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 / 0.03694) × 100

yield = 52.23 %

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5 0

3 years ago

What is the reaction type 2 Mgl2+ Mn(SO3)2 → 2 MgSO3 + Mnl4