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2 years ago

14

After mg ribbon is reacted completely with air, what is the reason to add some water and heat?

1 answer:

Vikentia [17]2 years ago

3 0

The reason why some water and heat are added after Mg ribbon is reacted completely with air is to remove Mg3N2 as MgO.

Combustion reaction is a reaction in which a substance is completely burnt in oxygen. Magnesium ribbon can be converted to magnesium oxide by combustion in air.

The reaction is as follows; 2Mg(s) + O2 ----> 2MgO(s)

Since air contains nitrogen, Mg3N2 is also formed. Heat and water can be added to the reaction thereby converting magnesium nitride to magnesium oxide.

Learn more: brainly.com/question/11527546

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If 1 kg of fuel is used in the above fusion reaction (2 1H + 3 1H--> 4 2He+1 0N) , the resulting helium has a mass of 0.993 k

enot [183]

Δmc
2

For one reaction:
Mass Defect =Δm
=2(m
H

)−m
He

−m
n

=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2

Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J

For 1 kg of Deuterium available,
moles=
2g
1000g

=500
N=500N
A

=3.01×10
26

Energy released =
2
N

×5.95×10
−13
J
=8.95×10
13

6 0

2 years ago

What is the pH for a solution that has an OH ion

Basile [38]

Assuming that you mean 10^-4 M then this would be basic and would have a pH of 10.

pOH = -log[OH].
So pOH = 4
pH=14-pOH
pH = 10

3 0

2 years ago

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod

Ipatiy [6.2K]

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂× $\frac{2 mol KClO_{3}}{3 mol O_{2}}$ = <em>0,15 mol of KClO₃</em>

I hope it helps!

8 0

3 years ago

The number of moles of molecules in a 12.0-gram sample of Cl2 is equal too?

Anvisha [2.4K]

it is equal theres your answer np :)

7 0

3 years ago

Read 2 more answers