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ipn [44]
3 years ago
8

How many minutes are there in 1.000 weeks

Chemistry
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

10080

Explanation:

You might be interested in
A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958
Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol  

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
Read 2 more answers
What volume is occupied by 0.104 mol of helium gas at a pressure of 0.91 atm and a temperature of 314 K ?
Aleksandr-060686 [28]

Answer:

The volume will be "2.95 L".

Explanation:

Given:

n = 0.104

p = 0.91 atm

T = 314 K

Now,

The Volume (V) will be:

= \frac{nRT}{P}

By putting the values, we get

= \frac{0.104\times 0.0821\times 314}{0.91}

= \frac{2.6810}{0.91}

= 2.95 \ L

7 0
2 years ago
An substance that is being dissolved by another object?​
Alex

Answer:

Solute - The solute is the substance that is being dissolved by another substance. In the example above, the salt is the solute. Solvent - The solvent is the substance that dissolves the other substance.

Explanation:

8 0
2 years ago
1. Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O
vovikov84 [41]
1) Balanced equation

C3H8 + 5O2 -> 3 CO2 + 4 H2O

2) 0.700 L C3H8

Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio

1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2

0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8

x = 3.500 L O2

3) CO2 produced

1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>

x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2

4) Water vapor produced

1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>

x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
3 0
3 years ago
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