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2 years ago

Quick electron emissions are called

Chemistry

1 answer:

vivado [14]2 years ago

5 0

Answer:

Beta decay

Quick electron emissions are called beta decay

Hope this helps

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I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

Question 2 (0 points)

bazaltina [42]

Answer:

The probability of spinning red with the spinner or rolling an odd number with the die is $\dfrac{1}{8}$

Explanation:

Given that,

Total color in spinner = 4

Let the area of four parts is equal in the spinner.

We need to calculate the probability of spinning red

Using formula of probability

$P(A) = \dfrac{Number\ of\ red\ color\ outcome}{Total\ number\ of\ colors\ in\ the\ spinner}$

Put the value into the formula

$P(A)=\dfrac{1}{4}$

We need to calculate the probability of odd number of the die

Using formula of probability

$P(B) = \dfrac{Number\ of\ odd\ digits\ outcome}{Total\ number\ of\ digits\ in\ the\ die}$

Put the value into the formula

$P(B)=\dfrac{3}{6}$

$P(B)=\dfrac{1}{2}$

We need to calculate the probability of spinning red with the spinner or rolling an odd number with the die

Using formula of probability of two events which is independent

$P(A\ and\ B)=P(A)\times P(B)$

Put the value into the formula

$P(A\ and\ B)=\dfrac{1}{4}\times\dfrac{1}{2}$

$P(A\ and\ B)=\dfrac{1}{8}$

Hence, The probability of spinning red with the spinner or rolling an odd number with the die is $\dfrac{1}{8}$

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3 years ago

Determine the molar mass of CaO

Umnica [9.8K]

56.0774 g/mol just took the test

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3 years ago

How many kJ are 3,340J?

forsale [732]

The answer I believe is 3.340kj.

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3 years ago

How many protons, and neutrons are in nuclide tin-115?

HACTEHA [7]

50 protons
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3 years ago

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