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nekit [7.7K]
3 years ago
14

Help me please!

Chemistry
1 answer:
Bumek [7]3 years ago
8 0
Good luck kid, now mark me brainliest
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Which of the following air pollutants causes a reddish-brown haze in the sky? Sulfur dioxide Carbon monoxide Nitrogen dioxide Ca
Zanzabum

The answer would most likely he sulfur dioxide

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3 years ago
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How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution r
NNADVOKAT [17]
2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃

m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol

1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw

2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}

m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}

m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}

m(Fe)=55.847*245*0.805/179.857= 61.24 g


3 0
3 years ago
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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0
3 years ago
What is the mass in grams of KBr in 0.400 L of a 0.350 M solution???
melamori03 [73]

Answer:

The answer to your question is: 16.7 g of KBr

Explanation:

Data

mass KBr = ? g

Volume = 0.400 L

Concentration = 0.350 M

Formula

Molarity = moles / volume

moles = molarity x volume

Process

moles = (0.350)(0.400)

          = 0.14

MW KBr = 39 + 80 = 119 g

                                  119 g of KBr --------------------  1 mol

                                   x                 --------------------   0.14 mol

                                   x = (0.14 x 119) / 1

                                   x = 16.7 g of KBr

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For which of the following molecules are resonance structures necessary to describe the bonding satisfactory ?
masya89 [10]
The answers is E: PF3
4 0
3 years ago
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