Which option is an example of physical change?

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

If the mass of the baking soda before the reaction was 10 grams and the mass of the vinegar was 15 grams and all of the baking s

Masteriza [31]

Answer:

8.946g NaHCO₃,

14.475g H₂O,

1.025g CH₃COO⁻Na⁺,

0.500g CO₂.

Explanation:

Baking soda (NaHCO₃) reacts with vinegar (CH₃COOH 5%) producing CH₃COO⁻Na⁺, H₂O and CO₂, thus:

NaHCO₃ + CH₃COOH → CH₃COO⁻Na⁺ + H₂O + CO₂

10g of NaHCO₃ are:

10g × (1mol / 84g) = 0.119 moles NaHCO₃

Moles of CH₃COOH in 15g of vinegar are:

15g vinegar × 5% = 0.75g CH₃COOH × (1mol / 60g) = 0.0125moles CH₃ COOH

After reaction, moles of NaHCO₃ that remains are:

0.119mol - 0.0125mol = <em>0.1065 moles NaHCO₃</em>

In mass:

0.1065mol ₓ (84g / 1mol) = <em>8.946g NaHCO₃</em>

Acetic acid reacts completely producing 0.0125moles of CH₃COO⁻Na⁺, water and CO₂. In mass:

CH₃COO⁻Na⁺ = 0.0125mol ₓ (82.03g / 1mol) = <em>1.025g CH₃COO⁻Na⁺</em>

H₂O = 0.0125mol ₓ (18.01g / 1mol) = <em>0.225g H₂O</em>

CO₂ = 0.0125mol ₓ (40g / 1mol) = <em>0.500g CO₂</em>

Now, 95% of vinegar is water (Doesn't react), that is:

15g × 95% = 14.25g H₂O. Total mass of water is:

14.25g + 0.225g = <em>14.475g H₂O</em>

Total mass is:

8.946g + 14.475g H₂O + 1.025g + 0.500g ≈ 25g. The same than initial mass following law of conservation of matter.

6 0

3 years ago

2KCIO3 -2KCL + 302

Firlakuza [10]

Answer:

a. 6 moles of O₂

b. 4.06×10²⁴ molecules of O₂

c. 3.04 g of KCl

Explanation:

Reaction of decomposition is:

2KClO₃ → 2KCl + 3O₂

a. See stoichiometry value

2 moles of potassium chlorate can decompose to 2 moles of potassium chloride and 3 moles of oxygen. Ratio is 2:3

If 2 moles of KClO₃ can decompose to 3 moles of O₂

Then 4 moles, may decompose to (4 . 3)/2 = 6 moles of O₂

b. In this case, the stoichiometry is the same.

Per 2 moles of KClO₃, I produce 2 moles of KCl

Then, 4.5 moles of KCl, were produced by 4.5 moles of KClO₃

We apply, the last relation:

(4.5 . 3) /2 = 6.75 moles of O₂ are also produced.

How many molecules are in 6.75 moles?

6.75 mol . 6.02×10²³ molecules/mol = 4.06×10²⁴ molecules of O₂

c. First of all, we convert the mass to moles:

5g . 1mol /122.55g = 0.0408 moles of salt

As ratio is 2:2, 0.0408 moles of salt, decompose to 0.0408 moles of KCl

We convert the moles to mass: 0.0408 mol . 74.55g /mol = 3.04g

5 0

2 years ago

Cho hai điện tích q1 = 2.10-6 C, q2 = 4. 10-6C, đặt tại A và B trong không khí biết AB = 2 cm. Xác định vectơ cường độ điện trườ

Nitella [24]

Answer:

Explanation:

<h2>sorry try lang baka mali</h2><h2 />

6 0

2 years ago

In which species does nitrogen have an oxidation number of zero?

VARVARA [1.3K]

The most common species of nitrogen in which its oxidation state is zero is the diatomic nitrogen ( $N_{2}$ ). This is because the oxidation number of pure elements (whether alone as an atom or combined with other atoms of the same element such as the diatomic nitrogen) is always zero.

In addition to diatomic nitrogen, there are diazonium ( $R-N_{2}^{+}$ ) compounds where nitrogen has an oxidation number of zero.

4 0

3 years ago