The reactants of two chemical equations are listed.

Which of these is an example of a wetland area? A. marsh B. stream C. city park D. farm land

Thepotemich [5.8K]

A.marsh it’s the most common one

5 0

3 years ago

Read 2 more answers

How many grams of H2Oare produced when 5.38g of O2<br> is reacted?

likoan [24]

Answer:

I think this answer should be 2.33 g H2O

6 0

2 years ago

Please help me with question 7. Thank you so much.

tigry1 [53]

Answer:

The system is not at equilibrium and the reaction will proceed to the left.

Explanation:

Step 1: Write the balanced equation

H₂(g) + CO₂(g) ⇄ CO(g) + H₂O(g)

Step 2: Calculate the reaction quotient (Q)

The reaction is calculated in the same way as the equilibrium constant (Kc) but it uses the concentrations at any time.

Q = [CO] × [H₂O] / [H₂] × [CO₂]

Q = 0.610 × 0.695 / 0.425 × 0.500 = 2.00

Since Q ≠ Kc, the reaction is not at equilibrium.

Since Q > Kc, the reaction will proceed to the left.

4 0

2 years ago

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical for

elena55 [62]

Answer:

(a) $C_5H_8O_2$

(b) $CHCl$

(c) $CH_2$

(d) $CH$

(e) $C_3H_3N$

Explanation:

Hello,

(a) For the lucite, one computes the moles of C, H and O that are present:

$n_C=0.599gC*\frac{1molC}{12gC}=0.05molC\\n_H=0.0806gH*\frac{1molH}{1gH}=0.0806molH\\n_O=0.32gO*\frac{1molO}{16gO}=0.02molO\\$

Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:

$C=\frac{0.05}{0.02} =2.5;H=\frac{0.0806}{0.02} =4;O=\frac{0.02}{0.02} =1$

Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:

$C_5H_8O_2$

(b) For the Saran, one computes the moles of C, H and Cl that are present:

$n_C=0.248gC*\frac{1molC}{12gC}=0.021molC\\n_H=0.02gH*\frac{1molH}{1gH}=0.02molH\\n_{Cl}=0.731gCl*\frac{1molCl}{35.45gCl}=0.021molCl\\$

Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:

$C=\frac{0.021}{0.02} =1;H=\frac{0.02}{0.02} =1;Cl=\frac{0.021}{0.02} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CHCl$

(c) For the polyethylene, one computes the moles of C and H that are present:

$n_C=0.86*\frac{1molC}{12gC}=0.072molC\\n_H=0.14gH*\frac{1molH}{1gH}=0.14molH$

Now, dividing each moles by the smallest moles (carbon's moles), one obtains:

$C=\frac{0.072}{0.072} =1;H=\frac{0.14}{0.072} =2$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH_2$

(d) For the polystyrene, one computes the moles of C and H that are present:

$n_C=0.923*\frac{1molC}{12gC}=0.077molC\\n_H=0.077gH*\frac{1molH}{1gH}=0.077molH$

Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:

$C=\frac{0.077}{0.077} =1;H=\frac{0.077}{0.077} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH$

(e) For the orlon, one computes the moles of C, H and N that are present:

$n_C=0.679*\frac{1molC}{12gC}=0.057molC\\n_H=0.057gH*\frac{1molH}{1gH}=0.057molH\\n_N=0.264gN*\frac{1molN}{14gN}=0.019molN$

Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:

$C=\frac{0.057}{0.019} =3;H=\frac{0.057}{0.019} =3;N=\frac{0.019}{0.019} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$C_3H_3N$

Best regards.

3 0

3 years ago

A solid and a liquid are shaken together in a test tube to produce a milky mixture that eventually separates into a clear liquid

mrs_skeptik [129]

This is called suspension

8 0

3 years ago