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mart [117]
2 years ago
6

2H2(g)+2NO(g)→2H2O(g)+N2(g) Part A If the concentration of NO changed from 0.100 M to 0.025 M in the first 15 minutes of the rea

ction, what is the average rate of the reaction during this time interval?
Chemistry
1 answer:
andrew11 [14]2 years ago
6 0

Answer:

0.0025 M/min

Explanation:

The rate of a reaction can be calculated for an element, based on its stoichiometric coefficient. For a reaction:

aA + bB = cC + dD , the rate will be

r = -(1/a)x(Δ[A]/Δt) = -(1/b)x(Δ[B]/Δt) = (1/c)x(Δ[C]/Δt) = (1/d)x(Δ[D]/Δt)

Where Δ[X] is the variation of the concentration of the X compound, Δt is the time variation, and the signal of minus in the reagents compounds is because they are disappearing, so Δ[X] will be negative, and r must be positive. So, for the reaction given:

r = -(1/2)x(Δ[NO]/Δt)

r = -(1/2)x( (0.025 - 0.1)/15)

r = 0.0025 M/min

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How many grams of barium sulfate, baso4, are produced if 25.34 ml of 0.113 m bacl2 completely react given the reaction: bacl2 +
dimulka [17.4K]
Answer is: mass of barium sulfate is 0.668 grams.
Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄<span> + 2NaCl.
V(</span>BaCl₂) = 25.34 mL ÷ 1000 mL/L = 0.02534 L.
c(BaCl₂) = 0.113 mol/L.
n(BaCl₂) = V(BaCl₂) · c(BaCl₂).
n(BaCl₂) = 0.02534 L · 0.113 mol/L.
n(BaCl₂) = 0.00286 mol.
From chemical reaction: n(BaCl₂) : n(BaSO₄) = 1 : 1.
n(BaSO₄) = 0.00286 mol.
m(BaSO₄) = n(BaSO₄) · M(BaSO₄).
m(BaSO₄) = 0.00286 mol · 233.4 g/mol.
m(BaSO₄) = 0.668 g.
8 0
3 years ago
Read 2 more answers
Which solid from the data table would float on water?
puteri [66]

Answer:

Ther answerr is A only solid A.

Explanation:

The density of water is 1 g/cm3 so anything above that will sink,  but as you can see solid A dosnt ev en reach one./ so It will float on water. the other two are over 1 g/cm3.

<h2>If this answer helped mark it brainliest. :)</h2>

5 0
3 years ago
The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of
tresset_1 [31]

Answer:

The total number of moles of gas in the mixture is 0.16939.

1.25550 grams of methane gas and 3.18848 grams of oxygen gas.

Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n =n_1+n_2

Using an ideal gas equation :

PV=nRT

P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K

P=2.509 n

Partial pressure of the methane= p_1=0.175 atm

Moles of the methane= n_1

Partial pressure of the oxygen gas= p_2=0.250 atm

Moles of the methane= n_2

Mole fraction  of the methane= \chi_1=\frac{n_1}{n_1+n_2}

Mole fraction  of the oxygen gas= \chi_2=\frac{n_2}{n_1+n_2}

p_1=p\times \chi_1  (Dalton's law)

0.175 atm=P\times \frac{n_1}{n_1+n_2}

0.175 atm=(2.509 n)\times \frac{n_1}{n}

n_1=0.06975 mol

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

p_2=p\times \chi_2  (Dalton's law)

0.250 atm=P\times \frac{n_2}{n_1+n_2}

0.250 atm=(2.509 n)\times \frac{n_2}{n}

n_2=0.09964 mol

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n =n_1+n_2

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]

7 0
3 years ago
2AlCl3 --&gt; 2Al + 3Al2
Eva8 [605]

Answer:

1597,5 gm AlCl3 decomposed

Explanation:

2AlCl3 --> 2Al + 3Al2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed? ​

FIRST, you need to write down the the correct reaction.  Your equation makes no sense

here is the correct question

2AlCl3 --> 2Al + 3Cl2    If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

for every mole of Al produced, we decomposed a mole of ​AlCl3

so we produced 15 moles of Al, so we decomposed 15 moles of AlCl3

15 moles are 15 X 106.5 = 1597,5 gm AlCl3 decomposed

7 0
2 years ago
Helppppppp plzzzzzzz
Alinara [238K]

Answer:

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Explanation:

3 0
2 years ago
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