What is the value for AG at 300 K if AH = 27 kJ/mol and AS = 0.09 kJ/(mol:K)? HURRY!

according to the law of conservation of , the number of atoms in the products of a chemical and equation must be balanced with t

Len [333]

yes because its kind of like a seesaw if theres too much on one side its unbalance there has to be the same amount of atoms on each side

5 0

3 years ago

What is the molarity of a solution prepared by dissolving 0.217 mol of ethanol, C2H5OH, in enough water to make 100.0 mL of solu

sergeinik [125]

Answer: 2.17mol/L

Explanation:

C=n/V

n: 0.217mol

V: 100.0mL

1. Convert mL to L (100.0mL/1000 = 0.01L)

2. Put numbers into equation (C= 0.217/0.01L)

3. Molarity is 2.17mol/L

8 0

3 years ago

Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2

Salsk061 [2.6K]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.7kJ$ ( × 2)

(2) $H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_2=-285.9kJ$ ( × 2)

(3) $2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_3=-1411kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 51.8 kJ.

6 0

3 years ago

What is a possible cause of a large percentage of error in an experiment where mgo is produced from the combustion of magnesium?

PilotLPTM [1.2K]

A possible cause of a large percentage of error in an experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has completely reacted. 

I hope this helps and if you have any further questions, please don’t hesitate to ask again.

4 0

3 years ago

Read 2 more answers

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Rufina [12.5K]

Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr

6 0

4 years ago