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2 years ago

If 250 g of calcium chlorate decomposes into calcium chloride and oxygen what mass of oxygen would be produced

Chemistry

1 answer:

Naya [18.7K]2 years ago

6 0

Answer:

115.936g

Explanation:

Ca(ClO3)2 -->CaCl2 +3 O2

One mole of calcium chlorate give up three moles of oxygen.

Molar mass of calcium chlorate is

40+(35.5+16*3)*2 = 207g/mol

Therefore there are

250g/207=

1.207mol of calcium cholrate...

So 1.207*3=3.623mol

Therefore 3.623mols of Oxygen must be produced.

Mass of the oxygen gas is equal to 3.623mol*32g/mol =115.936g

(O2 molar mass 32g/mol)

Cheers

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A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire

melamori03 [73]

Moles of Copper =

$1.52$ × $10^{22}atoms$ × $\frac{1 mol }{6.022 * 10^{23}atoms}$

= 0.0252 mol Cu

6 0

2 years ago

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass

inn [45]

Answer: The mass difference between the two is 7.38 grams.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

T = Temperature = $25^oC=[25+273]K=298K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For air:

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g$

Mass of air, $m_1$ = 8.56 g

For helium gas:

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g$

Mass of helium, $m_2$ = 1.18 g

Calculating the mass difference between the two:

$\Delta m=m_1-m_2$

$\Delta m=(8.56-1.18)g=7.38g$

Hence, the mass difference between the two is 7.38 grams.

5 0

2 years ago

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

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

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0

2 years ago

he first-order rate constant for the gas-phase decomposition of dimethyl ether, (CH3)2O → CH4 + H2 + CO is 3.2 ✕ 10−4 s−1 at 450

seropon [69]

Answer:

0.290 atm is the pressure of the system after 7.7min

Explanation:

The general first-order rate constant is:

ln [A] = -kt + ln [A]₀

Where [A] is concentration of A in time t,

K is rate constant, 3.2x10⁻⁴s⁻¹

[A]₀ is initial concentration = 0.336atm.

7.7 min are:

7.7min * (60s / 1min) = 462s

Solving:

ln [A] = -kt + ln [A]₀

ln [A] = -3.2x10⁻⁴s⁻¹*462s + ln [0.336atm]

ln [A] = -1.238

[A] =

<h3>0.290 atm is the pressure of the system after 7.7min</h3>

6 0

2 years ago

What is the density of a piece of metal in g/cm3 if its mass is determined to be 42.20 g and it is in the shape of a cube, with

maria [59]

It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3

density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.

density of object = 2.70 grams / cm^3 <<<<=== answer.

4 0

2 years ago

Read 2 more answers