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marishachu [46]
2 years ago
8

Calculate the kinetic energy of a baseball (mass = 5.25 oz) with a velocity of 1.0 x 10^2 mi/h

Chemistry
1 answer:
pychu [463]2 years ago
5 0

Answer:

262.5 Joules

Explanation:

You find the kinetic energy of multiplying half of the mass by the velocity. In this word problem it tells you the mass so you divide it by 2. That answer is 2.625, you then multiply that by the velocity, in this instance it's 1.0 x 10^2 mi/h.

The unit in kinetic energy is Joules. This is actually a really important part in chemistry and physics.

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In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0
3 years ago
Upon balancing the equation below, how many moles of sulfuric acid are needed to react completely with 8.4 moles of lithium hydr
maks197457 [2]
The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
7 0
3 years ago
How many milligrams are in 2.89 x 1024 CO2 molecules?
gogolik [260]

Answer:

2,959.36

Explanation:

7 0
3 years ago
Question 1
exis [7]

Answer: False

Explanation: adding heat to matter results in thermal energy.

4 0
3 years ago
What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?
GalinKa [24]
Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH

Now, we apply the definition of molality to calculate the molality of the solution:

mNaOH= \frac{moles NaOH}{kg_{solvent}}=  \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m

So, the molality of this solution is 4,41 m

Have a nice day!
4 0
3 years ago
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