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3 years ago

Coach Scott hits a baseball with his bat to help his players with fielding

Chemistry

1 answer:

dangina [55]3 years ago

8 0

Answer:

The ball exerts an equal and opposite force on the bat. This is the reaction force. Such an interaction pair is another example of Newton's Third Law. The baseball forces the bat in one direction and the bat forces the ball in the opposite direction.

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The electrons which are primarily responsible for chemical bonds are:

earnstyle [38]

Is there any choices because i can answer if you give me choices

4 0

3 years ago

Read 2 more answers

Determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. the mol

Softa [21]

To determine molecular formula, we first need to find out its empirical formula,
Carbon. Hydrogen. Nitrogen. Oxygen
Mass. 49.98g. 5.19g. 28.85g. 16.48g
Mole. 4.165. 5.19. 2.06. 1.03
Divide 4. 5. 2. 1
by
smallest
So by comparing the mole ratio from the table above, i hope u understand the table
The empirical formula is C4H5N2O
given molecular mass = 194.19g
so
(C4H5N2O) n= 194.19
(48+5+28+16)n=194.19
n= 2
molecular formula = C8H10N4O2

7 0

3 years ago

If 7.35 L of gas at 17°C and 3.55 atm is compressed at a pressure of 123 atm and 34°C, calculate the

MaRussiya [10]

Answer: V2= 0.224L

Explanation:

P1=3.55atm, V2= 7.35L, T1= 290, P2= 123atm, V2=?, T2= 307K

Applying general gas equation

P1V1/T1= P2V2/T2

Substitute and Simplify

3.55×7.35/290 = 123×V2/307

V2= 0.224L

6 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

HELP with this question!!!!!URGENT NEED ANSWER ASAP!!!!!! which of the following is true regarding single-displacement reactions

Tanya [424]

<span>Two reactants produce two products. </span>

7 0

3 years ago