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ycow [4]
2 years ago
5

Thor pushes a car with a mass of 2500 kg. It accelerates 5 m/s2. How much force did Thor apply?

Chemistry
1 answer:
ohaa [14]2 years ago
5 0

Answer:

<h3>The answer is 12500 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 2500 × 5

We have the final answer as

<h3>12500 N</h3>

Hope this helps you

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Oduvanchick [21]

Answer:

Dylan invested $ 600 in a savings account at a 1.6 % annual interest rate . He made no deposits or withdrawals on the account for 2 years. The interest was compounded annually . Find, to the nearest cent, the balance in the account after 2 years. a₁ What information from the question is important ? b . How much does Dylan have in his bank account after 2 years ? Show work .

4 0
2 years ago
Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper
KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

  • CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0
3 years ago
How many grams of carbonic acid were produced by the 3.00 g sample of NaHCO
stellarik [79]
I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
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mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3
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Ierofanga [76]

Explanation:

Percentage Yield

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