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Marianna [84]
3 years ago
6

The heat transfer of conduction is ...... than the heat transfer of convection

Chemistry
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

nnnnn

Explanation:

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Which element has the smallest atomic radius? Fluorine, Iron, Phosphorous, Potassium
Ivahew [28]

Answer:

actually answer should be helium

7 0
3 years ago
Read 2 more answers
State two differences between physical change and chemical change
Lubov Fominskaja [6]

Answer:

See explanation

Explanation:

1) Physical change is usually reversible, while chemical change isn't

2) Chemical change involves the change of chemical composition of matter while physical change doesn't

6 0
3 years ago
Read 2 more answers
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?
marin [14]
Balance Chemical Equation is as follow,

<span>                        Cu + 2 AgNO</span>₃     →    2 Ag + Cu(NO₃)₂

According to Balance Equation,

                   2 Moles of Ag is produced by reacting  =  1 Mole of Cu
So, 
     0.854 Moles of Ag will be produced by reacting  =  X Moles of Cu

Solving for X,
                             X  =  (0.854 mol × 1 mol) ÷ 2 mol

                             X  =  0.427 Moles of Cu
Result:
            0.854 Moles of Ag 
are produced by reacting 0.427 Moles of Cu.
4 0
3 years ago
Given the following UNBALANCED reaction: NH3 (g) &lt;--&gt; N2 (g) + H2 (g) If 1
Yakvenalex [24]

Answer:

C. 1.35

Explanation:

                                                     2NH3 (g) <-->          N2 (g) +             3H2 (g)

Initial concentration                2.2 mol/0.95L       1.1 mol/0.95L           0

change in concentration        2x                             x                           3x

                                                 -0.84 M                  +0.42M                +1.26M

Equilibrium                       1.4 mol/0.95L=1.47M        1.58 M                   1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M

8 0
3 years ago
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