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seropon [69]
3 years ago
13

Is barium peroxide Ba2O2 or BaO2?

Chemistry
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
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