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djverab [1.8K]
3 years ago
6

What are some chemacial changes in your school?

Chemistry
1 answer:
Oksanka [162]3 years ago
6 0
Putting glue on something, because once it is set in you cannot change it back.
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What is the energy transformation in the following:
Tatiana [17]
They all have something to do with electricity
3 0
3 years ago
A mixture of gases has 0.3000 mol of CO2, 0.2706 mol of SO2 and 0.3500 mol of water vapor. The total pressure of the mixture is
notka56 [123]

Answer:

Partial pressure SO₂ → 0.440 atm

Explanation:

We apply the mole fraction concept to solve this:

Moles of gas / Total moles = Partial pressure of the gas / Total pressure

Total moles = 0.3 moles of CO₂ + 0.2706 moles of SO₂ + 0.35 moles H₂O

Total moles = 0.9206 moles

Mole fraction SO₂ = 0.2706 moles / 0.9206 moles → 0.29

Now, we can know the partial pressure:

0.29 = Partial pressure SO₂ / Total pressure

0.29 = Partial pressure SO₂ / 1.5 atm

0.29 . 1.5atm = Partial pressure SO₂ → 0.440 atm

8 0
3 years ago
If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?
meriva
Use the clapeyron equation:

T in kelvin : 6.80 + 273 => 279.8 K

R = 0.082 

n = 71.5 moles

P = 5.03 atm

Therefore:

P x V = n x R x T

5.03 x V = 71.5 x 0.082 x 279.8

5.03 x V = 1640.4674

V = 1640.4674 / 5.03

V = 326.13 L

hope ths helps!
5 0
3 years ago
Fill in the blanks with appropriate options given in below:
jarptica [38.1K]

Answer:

1) -COOH

2) -NH2

3) hydrogen bonds

4) dispersion forces

5) -CH3

6) hydrogen bonds

7) negative

8) negative

9) positive

Explanation:

Alanine has a <u>-COOH</u> and a <u>-NH2</u> group available to form <u>hydrogen bonds</u> with water molecules.

Although there are some potential <u>dispersion forces</u> between the terminal <u>-CH3</u> group of alanine and hexane molecules, we expect the <u>hydrogen bonds</u> between alanine and water to be stronger.

Stronger intermolecular attractive forces between alanine and water lead to a more <u>negative ΔHmix</u> and more <u>negative (smaller positive)</u> ΔHsoln for water than for hexane.

4 0
3 years ago
If the atmospheric pressure in the laboratory is 1.2 atm, how many moles of gas were in each syringe? (Hint: Choose one volume a
Naily [24]

Answer:

A: 2.525 x 10-4 mol

B: 2.583 x 10-4 mol

Explanation:

Part A:

Data Given:

. Temperature of water (H2O) = 21.3°C

Convert Temperature to Kelvin

T = °C + 273

T = 21.3 + 273 = 294.3 K

volume of (H2O) gaseous state = 5.1 mL

Convert mL to liter

1000 mL = 1L

5.1 ml = 5.1/1000 = 0.0051 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n=PV/RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/ mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0051 L / 0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0061 atm.L / 24.162 L.atm.mol-1

n = 2.525 x 10-4 mol

no. of moles of gas (H2O) = 2.525 x 10-4 mol

Part B:

Data Given:

Temperature of water (H2) = 21.3°C

Convert Temperature to Kelvin

T = "C + 273

T= 21.3 + 273 = 294.3 K

volume of (H2) gas = 5.2 mL

Convert mL to liter

1000 mL = 1 L

5.2 ml = 5.2/1000 = 0.0052 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n= PV / RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0052 L/0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0062 atm.L/ 24.162 L.atm.mol-1

n = 2.583 x 10-4 mol

I

no. of moles of gas (H2) = 2.583 x 10-4 mol

8 0
3 years ago
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