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2 years ago

Review the equation below.

Chemistry

1 answer:

rusak2 [61]2 years ago

3 0

Answer:

C) 3

Explanation:

Given data;

Number of moles of oxygen produced = ?

Moles of KClO₃ decomposed = 2 mol

Solution:

Chemical equation;

2KClO₃ → 2KCl + 3O₂

Now we will compare the moles of KClO₃ with O₂

KClO₃ : O₂

2 : 3

Thus, 3 moles of oxygen are produced.

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Two different compounds are formed by the elements nitrogen and oxygen. The first compound, compound P1, contains 64.17% by mass

xz_007 [3.2K]

Answer:

Explanation:

To solve the problem, we must know the kind of compounds we are dealing with.

For the first compound, P1 and second compound P2:

N O N O

Mass percent 64.17 35.73 47.23 52.79

Atomic mass 14 16 14 16

Number of

moles 64.17/14 35.73/16 47.23/14 52.79/16

4.58 2.23 3.37 3.30

Simplest

ratio 4.58/2.23 2.23/2.23 3.37/3.30 3.3/3.3

2 1 1 1

P1 compound is N₂O

P2 compound is NO

These are the compounds,

In N₂O = 28:16

NO = 14:16

This is the ratio of nitrogen to a fixed mass of oxygen for the two compounds.

7 0

2 years ago

A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

2 years ago

During photosynthesis, plants use carbon dioxide, water, and light energy to produce glucose, C

Akimi4 [234]

The outcome of the equation shows that there is no light energy proving that light energy was absorbed to get CH20; +602

3 0

3 years ago

Anybody wanna do a zoom

Ksivusya [100]

For what exactly ? just for fun or for educational purposes?

8 0

2 years ago

How much water can be raised from 25*C (room temperature) to 37*C (body temperature) by adding the 2,000 kJ in a Snickers bar?

yuradex [85]

Answer:

m = 39834.3 g

Explanation:

Given data:

Mass of water raised = ?

Initial temperature = 25°C

Final temperature = 37°C

Energy added = 2000 Kj (2000 ×1000= 2000,000 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 37°C - 25°C

ΔT = 12°C

c = 4.184 g/j.°C

Q = m.c. ΔT

2000,000j = m .4.184 g/j.°C. 12°C

2000,000j = m. 50.208 g/j

m = 2000,000j / 50.208 g/j

m = 39834.3 g

6 0

2 years ago