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3 years ago

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Is barium peroxide Ba2O2 or BaO2?

2 answers:

vredina [299]3 years ago

6 0

Answer:options?

Explanation:

astraxan [27]3 years ago

6 0

Barium peroxide is BaO2

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Question14 of 20If 5.15 g FeCl3 is dissolved in enough water to make exactly 150.0 mL of solution, what is the molar concentrati

Serhud [2]

Answer: 0.635 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= Moles= $\frac{\text{Given mass}}{\text{Molar mass}}=\frac{5.15g}{162.4g/mol}=0.0317moles$

$V_s$ = volume of solution = 150 ml

$Molarity=\frac{0.0317\times 1000}{150ml}=0.2114M$

$FeCl_3\rightarrow Fe^{3+}+3Cl^-$

as 1 mole of $FeCl_3$ gives 3 moles of $Cl^-$ ions

Thus molarity of $Cl^-$ = $3\times 0.2114=0.635M$

Molarity of $Cl^-$ = 0.635 M

8 0

3 years ago

What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?

MArishka [77]

Combined gas law is
PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K

By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL

5 0

3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago

Jim draws a wave that looks like this wave:

Travka [436]

wave length is what he decreased

8 0

3 years ago

Read 2 more answers

2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is r

vredina [299]

Answer :

(a) The rate of $NO_2$ formed is, 0.066 M/s

(b) The rate of $O_2$ formed is, 0.033 M/s

Explanation : Given,

$\frac{d[NO]}{dt}$ = 0.066 M/s

The balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $O_2$ = $-\frac{d[O_2]}{dt}$

The rate of formation of $NO_2$ = $\frac{1}{2}\frac{d[NO_2]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ = 0.066 M/s

(a) Now we have to determine the rate of $NO_2$ formed.

$\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s$

The rate of $NO_2$ formed is, 0.066 M/s

(b) Now we have to determine the rate of molecular oxygen reacting.

$-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

The rate of $O_2$ formed is, 0.033 M/s

6 0

3 years ago