Why do chemical equations need to be balanced?

A gas container has a volume of 446.9 with a temp of 14c. When the volume is decreased to 238.7l the new temp is what

swat32

Answer:

$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2} } \\ \frac{446.9}{(14 + 273)} = \frac{238.7}{T _{2} } \\ {T _{2}} = \frac{238.7 \times 287}{446.9} \\ {T _{2}} = 153.3 \: kelvin \\ = 119.7 \degree \: c$

7 0

3 years ago

Which statement best describes temperature? A.

USPshnik [31]

?? post the statements

4 0

4 years ago

Read 2 more answers

Please help me i will give brain list

sesenic [268]

Answer:

a. pentane = C5H12

b. Heptane = C7H16

c. Hexyne = C6H10

d. Octene = C8H16

e. Nonane = C9H20

Hope it helps.

7 0

3 years ago

Read 2 more answers

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0

3 years ago

1.What is the specific heat capacity of granite when 20 kg absorbs 237 000 J of heat energy, causing its temperature to increase

Natalka [10]

Given :

Amount = 20 kg
Heat energy absorbed = 237,000 J
Temperature change = 15 °C

Formula applied :

$\boxed {Q = mc \triangle T}$

Q = absorbed heat
m = mass
c = specific heat capacity
ΔT = temperature change

Let's solve for c !

⇒ 237,000 = 20 × c × 15

⇒ c = 237,000 ÷ 300

⇒ $\boxed {c = 790 J kg^{-1} K^{-1}}$

∴ The specific heat capacity of granite is 790 J kg⁻¹ K⁻¹.

5 0

2 years ago