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3 years ago

What is a submatomic particle

Chemistry

1 answer:

Amanda [17]3 years ago

5 0

Answer: A subatomic particle is a particle smaller than an atom, hence the name broken down into a sub category

Explanation:

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Which statements describe Rutherford’s model of the atom? Select three options.

AnnZ [28]

Answer:

Well, I cannot see the options but if I were you I would choose the one closest to this. Rutherford's model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.

Explanation:

Again I cannot see the options but here is what I would guess. Hope this helped and have a great day! :-)

6 0

2 years ago

Read 2 more answers

A 35.0 mL sample of 1.00 M KBr and a 60.0 mL sample of 0.600 M KBr are mixed. The solution is then heated to evaporate water unt

Katarina [22]

Answer: The molarity of KBr in the final solution is 1.42M

Explanation:

We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.

We will first calculate the number of moles of KBr in the individual sample before mixing together

In the first sample:

Volume (V) = 35.0 mL

Concentration (C) = 1.00M

Number of moles (n) = C × V

n = (35.0mL × 1.00M)

n= 35.0mmol

For the second sample

V = 60.0 mL

C = 0.600 M

n = (60.0 mL × 0.600 M)

n = 36.0mmol

Therefore, we have (35.0 + 36.0)mmol in the final solution

Number of moles of KBr in final solution (n) = 71.0mmol

Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.

Therefore,

Final volume of solution (V) = 50mL

Number of moles of KBr in final solution (n) = 71.0mmol

From

C = n / V

C= 71.0mmol/50mL

C = 1.42M

Therefore, the molarity of KBr in the final solution is 1.42M

5 0

2 years ago

The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

hjlf

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

$K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm$

= 4.7 × 10⁻⁴atm

4 0

3 years ago

Read 2 more answers

For every action, there is an equal and opposite reaction. true or false

Svetlanka [38]

True because evryone reacts a different way

5 0

2 years ago

Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0

3 years ago