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anastassius [24]
3 years ago
10

Na + MgF^2 = NaF+ Mg

Chemistry
1 answer:
raketka [301]3 years ago
5 0

Answer:

HAAAAA U IN SOME DEEP S***

Explanation:

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What is the best way to package and preserve a questioned document?
Zanzabum
In a plastic sheet protector
5 0
3 years ago
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1. What are the possible genotypes of offspring when crossing BB, Bb parents?
marissa [1.9K]

Answer:

1, C: BB, Bb, Bb, BB

2. C: Hybrid

Explanation:

1. If u do a punnet square for BB and Bb you will get: BB, Bb, Bb, Bb

B| B|

B| BB. BB

b| Bb Bb

2. You do a punnet square for BB and bb and you'll get: Bb, Bb, Bb, Bb, which means all the children are hybrids of Dominant alleles ans recessive alleles

<u>B </u> <u>B</u>

b| Bb Bb

b| Bb Bb

4 0
2 years ago
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The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

4 0
3 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
Name some acids and in what and how we used them
Ne4ueva [31]

Answer:

Acid Uses

Organic acids

Citric acid As a preservative for food As a flavouring agent

Ascorbic acid (also called vitamin C) In the treatment of bone marrow and scurvy diseases

Acetic acid Added to pickles to make them sour

Explanation:

7 0
2 years ago
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