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harina [27]
4 years ago
15

When heated KClO3 decomposes into KCl and O2 if this reaction produced 31.9 g of KCl how much O2 was produced in grams

Chemistry
1 answer:
deff fn [24]4 years ago
3 0
I think the answer will be 13.1 g O2
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Explanation:

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What are the products of the combustion of a hydrocarbon?
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Answer:

I think carbon and hydrogen

7 0
3 years ago
Read 2 more answers
Balancing Redox Reactions<br> VO2++MnO4 V(OH)4++Mn2+(acidic)
Natasha_Volkova [10]

Answer:

MnO4- + 5 VO +2 + 11 H2O = 5 V(OH)4+ + Mn+2 + 2 H+

Explanation:

3 0
3 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

a

a

∣

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0
4 years ago
C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
3 years ago
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