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3 years ago

40 IF YOU ANSWER CORRECTLY & ADDED BRAINIEST

Chemistry

1 answer:

Vikentia [17]3 years ago

6 0

Answer:

1. B

2. C

Explanation:

1. There are 15 electrons which means that the element is phosphorus. We know it is in the excited state because an electron from the second ring has moved to the third one, typically it would be 2 - 8 - 5.

2. An element with 3 valence electrons will be 3 spaces from the left side of the periodic table. Sodium is 1, Magnesium is 2, Aluminium is 3, and Silicon is 4. Noble gasses on the right have all 8.

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A force of 20 N acts upon 5 kg block caculate the accerleration of the object

Vedmedyk [2.9K]

Answer:

$\boxed{\text{4 m $\cdot$ s$^{-2}$}}$

Explanation:

F = ma

$a = \dfrac{F}{m} = \dfrac{\text{20 N}}{\text{5 kg}} \times \dfrac{\text{1 kg$\cdot$ m $\cdot$ s$^{-2}$}}{\text{1 N}} = \text{4 m $\cdot$ s$^{-2}$}\\\\\text{The acceleration of the object is } \boxed{\textbf{4 m $\cdot$ s$^{\mathbf{-2}}$}}$

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3 years ago

Plz HELP ITS SCIENCE!

Alexxx [7]

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6 0

3 years ago

Read 2 more answers

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: pcl3(g+cl2(g?pcl5(g. a 7.5-l gas vessel

Natasha2012 [34]

25655+6565++65+65+65+56+566+56+556+5+656+56+56+56+56+

6 0

4 years ago

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e

hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

CxHyOz + O₂ ⇒ CO₂ + H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

44 g of CO₂ -------------- 12 g of C

3.268 g of CO₂ -------- x

x = (3.268 x 12) / 44

x = 0.891 g of Carbon

12 g of carbon ----------- 1 mol

0.891 g of C ---------- x

x = (0.891 x 1) / 12

x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

18 g of water --------------- 2 g of H

1.672 g of H₂O ------------ x

x = (1.672 x 2) / 18

x = 0.186 g of hydrogen

1 g of hydrogen ------------ 1 mol of H

0.186 g of H ------------ x

x = (0.186 x 1) / 1

x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

= 0.299 g of O₂

Moles of Oxygen

16 g of Oxygen ---------------- 1 mol

0.299 g of O ----------------- x

x = (0.299 x 1) / 16

x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon 0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen 0.186/ 0.019 = 9.7 = 10

Oxygen 0.019/ 0.019 = 1

5.- Write the empirical formula

C₄H₁₀O

4 0

4 years ago

Consider the reaction given below.

Drupady [299]

Answer:

K = 0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.

Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

2ᵃ = 2 ⇒ a = 1

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K = 0.167 s⁻¹

6 0

4 years ago

*40 IF YOU ANSWER CORRECTLY & ADDED BRAINIEST*

40 IF YOU ANSWER CORRECTLY & ADDED BRAINIEST