What is the formula for volume of a rectangular solid?

Chemistry

2 answers:

Vladimir [108]3 years ago

7 0

Answer:

1st.

thats the one for u to choose

WINSTONCH [101]3 years ago

5 0

Answer:

1. length x width x height

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Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)

klio [65]

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

Best regards.

8 0

3 years ago

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:

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