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STALIN [3.7K]
3 years ago
12

CORRECT ANSWER GETS BRAINLIEST!!!!!!!!!!!!

Chemistry
2 answers:
Mrac [35]3 years ago
8 0

Answer:

2, 4, 1, 3, 5

Explanation:

i believe is this a test

please mark this answer as brainlest i got the answer from someone by the way which is my sis

but forget all of that

please mark brainlest

Mama L [17]3 years ago
5 0
3,2,1,5,4 I’m pretty sure
You might be interested in
What is the standard gibbs free energy for the transformation of diamond to graphite at 298 k? cdiamondâcgraphite?
denis-greek [22]
Considering; graphite; standard enthalpy = 0 and entropy = 5.740; diamond standard enthalpy = 1.897 and entropy = 2.38.
Using the equation Delta G = Delta H - Temperature (DeltaS)
Delta H = enthalpy sum of products - enthalpy sum of reactants
Which will be; 0 -1.897 = -1.897 kJ/Mol
Delta S is the entropy sum; given by 
5.740 - 2.38 = 3.36 J/Mol
We can convert Delta S from Joules to kilo Joules by dividing by 1000
we get ; 0.00336 kJ/mol
We are given a temperature in kelvin which suits the calculations ((298 k)
Therefore; using the equation; 
= -1.897 - (298 × 0.00336) = -2.90 kJ
Thus; the standard gibbs free energy will be; -2.9 kJ

5 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
4 years ago
How many grams are in 4.63 x 1024<br><br> molecules of CCl4?
MrRa [10]

Answer:

m=1,182.8g

Explanation:

Hello!

In this case, since the relationship between molecules and mass is first analyzed via the Avogadro's number to compute the moles in the given molecules:

mol=4.63x10^{24}molec*\frac{1mol}{6.022x10^{23}molec}=7.69mol

We now use the molar mass of carbon tetrachloride (153.81 g/mol) to obtain the required grams:

m=7.69mol*\frac{153.81g}{1mol} \\\\m=1,182.8g

Best regards!

6 0
3 years ago
Calculate the atomic mass of element X if there are two isotopes of this element (X-24 and X-28) and 75% of all X atoms are X-28
Zina [86]
[(25% x 24) + (75% x 28)]/100 = 27 amu. The answer is d. 27 amu
5 0
3 years ago
Read 2 more answers
What is the maximum mass of glucose (C6H12O6) that can be burned in 10 g of oxygen?
Dahasolnce [82]
C_6H_{12}O_6 + 6 \ O_2 \to 6 \ CO_2 + 6 \ H_2O
1 mole of glucose : 6 moles of oxygen

First calculate the number of moles of oxygen in 10 g:
M=32 \ \frac{g}{mol} \\&#10;m=10 \ g \\&#10;n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol

1 mole of glucose reacts with 6 moles of oxygen
x moles of glucose reacts with 0.3125 moles of oxygen
x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol

Now calculate the mass of 5/96 moles of glucose.
M=180 \ \frac{g}{mol} \\&#10;n=\frac{5}{96} \ mol \\&#10;m=180 \ \frac{g}{mol} \cdot \frac{5}{96} \ mol=\frac{900}{96} \ g=9.375 \ g

The maximum mass of glucose that can be burned in 10 g of oxygen is 9.375 g.
5 0
3 years ago
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