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4 years ago

In a reaction, 25 grams of reactant

Chemistry

1 answer:

Vsevolod [243]4 years ago

8 0

Answer:

D. 15g

Explanation:

The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

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To the right of calcium and in the same period.

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3 years ago

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A solution is prepared at 25°C that is initially 0.18 M in methylamine (CH3NH2), a weak base with Kb = 4.4 x 10^-4, and 0.35 M i

nlexa [21]

Answer:

pH of the solution is 10.37

Explanation:

$pOH=pkb+log\frac{[salt]}{[base]}$

kb = $4.4 \times 10^{-4}$

pkb = -log kb

= $-log4.4 \times 10^{-4}$

= 3.35

salt is methylammonium bromide and methylamine is base

Substitute the values in the above expression as follows:

$pOH=pkb+log\frac{[salt]}{[base]} \\=3.35+log\frac{0.35}{0.18} \\=3.35+0.28\\=3.63$

pH = 14 - pOH

= 14 - 3.63

= 10.37

pH of the solution is 10.37

3 0

3 years ago

1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to<br> 15°C?

weqwewe [10]

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2= 15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

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3 years ago

A gas mixture being used to simulate the atmosphere of another planet at 23°c consists of 337 mg of methane, 148 mg of argon, an

Karolina [17]

The total pressure of the mixture is 65.5 kPa.

According to Dalton's Law of Partial Pressure,

The partial pressure of gas = Mole fraction of gas × Total pressure

Total Pressure = Sum of all the gases partial pressures

The number of moles of methane is,

$Moles \: of \: methane \: (16 g/mol) = 337 \: mg \times \frac{1 g}{1000 mg} \times \frac{ 1 mol}{16 g }$

= 0.021 mols

The moles of methane are 0.021 mols.

The number of moles of the argon,

$Moles \: of \: argon (40 g/mol) = 148 \: mg \times \frac{ 1 g}{1000 mg } \times \frac{ 1 mol}{40 g}$

= 0.003 mols

The number of moles of argon is 0.003 mols.

The number of moles of nitrogen is,

$Moles \: of \: nitrogen (28 g/mol) = 296 \: mg \times \frac{ 1 g}{1000 mg} \times \frac{ 1 mol/}{28 g}$

= 0.010 mols

The number of moles of nitrogen is 0.010 mols.

The total number of moles is,

= 0.021 + 0.003 + 0.010

= 0.034 mols

$Mole \: fraction = \frac{ Moles \: of \: solute }{Total \: number \: of \: moles \: of \: soulte \: and \: solvent}$

$= \frac{ 0.010 }{ 0.034}$

= 0.29

$0.29 \: P _{total} = 19 \: kPa$

$P _{total} = \frac{ 19 \: kPa }{0.29}$

= 65.5 kPa

Therefore, the total pressure of the mixture is 65.5 kPa.

To know more about Dalton's law, refer to the below link:

brainly.com/question/14119417

#SPJ4

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1 year ago

Jamal was riding his bike home at a speed of 12 mph. On his trip out, he had traveled 8 miles away from his house. He has been t

NikAS [45]

Jamal is 2 miles away from his home.

Option B.

<h3><u>Explanation:</u></h3>

The initial distance of Jamal from his home is 8 miles. He needs to return home from his trip.

The speed that Jamal kept was constant for 30minutes and he drove at a speed for 12 miles per hour.

Speed of Jamal = 12 mph.

Time of travel = 30minutes =0.5 hours.

So distance traveled by Jamal = $12 \times0.5$ miles. = 6miles.

Initial distance of Jamal = 8 miles.

So final distance = $8-6$ miles = 2miles.

So Jamal is 2 miles away from his home.

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3 years ago

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