A

What isotope name has a mass number of 201

Svetlanka [38]

Answer:

Thallium is an isotope name that has a mass number of 201

3 0

4 years ago

Beverly and Carl are in a race. Their graphs show the data.

Nikitich [7]

Given the data, the correct statement is

Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

<h3>How to determine the initial acceleration of Beverly</h3>

Initial velocity (u) = 0 m/s
Final velocity (v) = 15 m/s
Time (t) = 10 s
Initial acceleration (a₁) =?

a₁ = (v – u) / t

a₁ = (15 – 0) / 10

a₁ = 1.5 m/s²

<h3>How to determine the final acceleration of Beverly</h3>

Initial velocity (u) = 15 m/s
Final velocity (v) = 35 m/s
Time (t) = 50 - 10 = 40 s
Final acceleration (a₂) =?

a₂ = (v – u) / t

a₂ = (35 – 15) / 40

a₂ = 0.5 m/s²

<h3>How to determine the initial acceleration of Carl</h3>

Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 10 s
Initial acceleration (a₁) =?

a₁ = (v – u) / t

a₁ = (10 – 0) / 10

a₁ = 1 m/s²

<h3>How to determine the final acceleration of Carl</h3>

Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Time (t) = 50 - 10 = 40 s
Final acceleration (a₂) =?

a₂ = (v – u) / t

a₂ = (30 – 10) / 40

a₂ = 0.5 m/s²

SUMMARY

Initial acceleration of Beverly = 1.5 m/s²
Final acceleration of Beverly = 0.5 m/s²
Initial acceleration of Carl = 1 m/s²
Final acceleration of Carl = 0.5 m/s²

From the above calculations, we can see that Beverly's initial acceleration is higher than that of Carl's and their final acceleration is the same.

Therefore, the correct answer to the question is:

Even though for a majority of the race they accelerated at the same rate, Beverly won because her initial acceleration was greater than Carl’s

Complete question

See attached photo

Learn more about acceleration:

brainly.com/question/491732

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8 0

2 years ago

For the reaction where Δn=−1Δn=−1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactant

Alex787 [66]

Answer:

Explanation:

In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases.

Δn= b - a

Δn= moles of gaseous products - moles of gaseous reactants

Therefore, <u>after the increase in volume</u>:

If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.

8 0

3 years ago

A chemist heats a fixed amount of gas in a sealed glass container. Which law is the chemist most likely investigating?

fiasKO [112]

Answer:

c

Explanation:

was c the correct answer or not

5 0

3 years ago

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

7 0

3 years ago