Chemistry!! Please help asap!! Will mark brainiliest!!

Chemistry

1 answer:

Whitepunk [10]3 years ago

5 0

<h2>Answer:</h2><h3>Part 1. </h3>

Option B is correct option.

The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing OH- ions.

Explanation:

Full equation:

2MnO2 + H2O + 2e- → Mn2O3 + 2OH-

The option B which is Mg is stronger reducing agent than Ag is correct option.

Explanation:

Equation:

Mg(s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag(s)

According to equation Mg converts to Mg+2 which means it gives to electron to reduce Ag. So it act as an reducing agent.

The correct option is B. Which is 5, 1, 8, 5, 1, 4.

Explanation:

Full equation :

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) --> 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O (l)

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Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

3 years ago

A sample of the mineral hematite Iron (III) oxide has a mass of 12.4g. How many moles of the mineral are present?

8090 [49]

Thus problem is providing us with the mass of iron (III) oxide as 12.4 g so the moles are required and found to be 0.0776 mol after the calculations:

<h3>Mole-mass relationships:</h3>

In chemistry, we use mole-mass relationships in order to calculate grams from moles and vice versa. In this case, since we are given the mass of iron (III) oxide as 12.4 g one can calculate the moles by firstly quantifying its molar mass: