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2 years ago

A weather balloon contains 13.0 L of helium at a pressure of 1.01 atm and a temperature of 21.0°C. If this had

Chemistry

1 answer:

Stolb23 [73]2 years ago

5 0

Answer:

2.45atm

Explanation:

Use Combined Gas Law: $\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

Solve For P₂: $P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$

Use Kelvin: 21.0C = 294.15K 29.0 = 302.15K

(1.01atm)(13.0L)(302.15K)=3967.2295

(294.15K)(5.50L)=1617.825

3967.2295/1617.825=2.45219940352 = 2.45 atm

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A large sport utility vehicle has a mass of 2600 kg.

Pepsi [2]

Mass of CO2 = 225.632g

4 0

2 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago

Where is most of Earth's liquid water ? what must be done so humans can drink it ?

LiRa [457]

You're right it's the oceans.. since most is salt water, the salt has to be boiled out

4 0

2 years ago

Read 2 more answers

What is the IUPAC name for this product?

mihalych1998 [28]

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

In the given product total four carbon atoms are present and between all of them single bonds are present.
In the second carbon atom, chlorine group is present.
During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

To know more about IUPAC nomenclature, visit the below link:

brainly.com/question/26635784

#SP1

7 0

1 year ago

For double-helix formation, DG can be measured to be 2 54 kJ mol 1 ( 2 13 kcal mol 1 ) at pH 7.0 in 1 M NaCl at 25 8C (298 K). T

levacccp [35]

Answer:

ΔS = -661.0J/mol is the entropy change for the system

ΔS = -842J/mol.K is the entropy change for the surroundings

Explanation:

From the relationship between ΔG, T, ΔH and ΔS,

Mathematically, ΔG = ΔH - TΔS

TΔS = ΔH - ΔS

ΔS = ΔH - ΔS / T

but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)

plugging into the equation,

ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298

ΔS = -0.6610KJ/mol or in J.mol

ΔS = -661.0J/mol is the entropy change for the system

For entropy change for the surroundings = ΔS = ΔH/T

ΔS = -251 kJ/mol / 298K

ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings

8 0

2 years ago