Answer:
3.65 mol O₂
Explanation:
Step 1: RxN
2H₂O₂ → 2H₂O + O₂
Step 2: Define
Given - 7.30 mol H₂O₂
Solve - x mol O₂
Step 3: Stoichiometry
[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂
What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)
Answer:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Explanation:
Hello.
In this case, for the undergoing chemical reaction:
[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]
We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Best regards!
A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain
Answer:
[tex]Total = 50.6\ moles[/tex]
Explanation:
Given
[tex]Propane = C_3H_8[/tex]
Represent Carbon with C and Hydrogen with H
[tex]C = 13.8[/tex]
Required
Determine the total moles
First, we need to represent propane as a ratio
[tex]C_3H_8[/tex] implies
[tex]C:H = 3:8[/tex]
So, we're to first solve for H when [tex]C = 13.8[/tex]
Substitute 13.8 for C
[tex]13.8 : H = 3 : 8[/tex]
Convert to fraction
[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]
Cross Multiply
[tex]3 * H = 13.8 * 8[/tex]
[tex]3 H = 110.4[/tex]
Solve for H
[tex]H = 110.4/3[/tex]
[tex]H = 36.8[/tex]
So, when
[tex]C = 13.8[/tex]
[tex]H = 36.8[/tex]
[tex]Total = C + H[/tex]
[tex]Total = 13.8 + 36.8[/tex]
[tex]Total = 50.6\ moles[/tex]
When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation
Answer:
during sublimation
Explanation:
just took the test
A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.
a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g
Answer:
you can see the empirical formula at the pic
The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.
What is empirical formula?
Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.
(a) 1.245 g Ni : 5.381 g I
Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2
So the formula is NiI2
(b) 2.677 g Ba : 3.115 g Br
Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038
= 0.02 : 0.04 = 1:2
So the formula is BaBr2
(c) 2.128 g Be : 7.557 g S
Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1
So the formula is BeS
Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.
To learn more about empirical formula, refer to the link below:
https://brainly.com/question/11588623
#SPJ2
Which of the following is an Elementary compound?
A. CO2
B. N2
C. SO2
D. H2S
heeeeeeeeeelp please please please
Answer:
Explanation:
In my opinion the answer should be SO2
Answer:
a should be answer i think.