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Kobotan [32]
3 years ago
11

Elements in an element group have similar

Chemistry
1 answer:
svetlana [45]3 years ago
8 0

Answer:

chemical properties and the same number of valence electrons

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Which element has the following electron configuration: 1s²2s²2p⁶3s²3p⁶4s²3d104p⁶5s²4d105p⁶6s²5d¹4f145d⁶
likoan [24]
The element is Os (Osmium)
4 0
3 years ago
determine the percent yield for carbon dioxide if 4.50 moles of propane yielded 7.64 moles of carbon dioxide
ira [324]

Answer:

Percent yield = 57%

Explanation:

Given data:

Number of moles of propane = 4.50 mol

Number of moles of carbon dioxide = 7.64 mol

Percent yield = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Now we will compare the moles of propane and carbon dioxide.

                            C₃H₈            :            CO₂

                                 1               ;                3

                                  4.50        :              3×4.50 = 13.5 mol

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield =  7.64 mol / 13.5 mol × 100

Percent yield = 0.57× 100

Percent yield = 57%

3 0
3 years ago
Lonization energy is the
Tamiku [17]
Arrow on the table below draw an
3 0
3 years ago
How many moles of MgS2O3 are in 223 g of the compound
Alexxandr [17]
Moles of MgS2O3 = 223/molar mass of MgS2O3
    
                              =   223/136.42 
                              =     1.634 moles.

Hope this helps!
6 0
3 years ago
Read 2 more answers
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
4 years ago
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