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3 years ago

A softball is thrown with 145 Joules of kinetic energy. If the ball is moving at 20.0 m/s, what is the mass of the ball in kg?

Chemistry

1 answer:

Tasya [4]3 years ago

3 0

Answer:

0.725 kg

Explanation:

Step 1: Given data

Kinetic energy of the softball (K): 145 J

Speed of the softball (v): 20.0 m/s

Mass of the softball (m): ?

Step 2: Calculate the mass of the softball

We will use the following expression.

K = 1/2 × m × v²

m = 2 K / v²

m = 2 × 145 J / (20.0 m/s)²

m = 0.725 kg

The mass of the softball is 0.725 kg.

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3 years ago

A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.

n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) = (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1 = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1 - 1/ T3 = 1/400 - 1 /425 = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

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3 years ago

For the reaction 2 Cr(s) + 3 Pb²⁺(aq) ⟶ 3 Pb(s) + 2 Cr³⁺(aq), what is the value of n in the Nernst equation?

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Answer:

Explanation:

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3 years ago

Ammonium hydroxide is _____.<br> an acid<br> a base<br> a neutral

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Ammonium hydroxide is basic

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3 years ago

What is the average yearly rate of change of carbon-14 during the first 5000 years?

erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

$A = A_0 \times e^{-\lambda \times t}$

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

$T_{1/2}$ = The half life

$\lambda = 0.693/T_{1/2}$

t = The time elapsed = 5000 years

λ = 0.693/ $T_{1/2}$ = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0

3 years ago