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slega [8]
4 years ago
12

How are compounds with ionic and covalent bonds alike?

Chemistry
1 answer:
PSYCHO15rus [73]4 years ago
8 0

Answer:

B

Explanation:

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Select the fraction that is equivalent to 1/2
zhannawk [14.2K]

Answer:

5/10,

10/20,

3/6

4/8

6/12

3 0
3 years ago
7. Which statement is true about particles in ice and in boiling water?
IceJOKER [234]

Answer:

A. particles are the same.

Explanation:

Hello, since we are talking about the water, the molecules are quite equal.

Nonetheless,  if you want to know it (because this is not in the answers), the particles move faster in the boiling water than in the ice, considering that in the boiling water there is a larger energy content, which allows the molecules to move faster (they "have" more energy). In addition, The molecules are closer in the ice because that is the only way for them to solidify (get together).

Best regards!

6 0
3 years ago
When 8.0 g H₂ react with 8.0 g O₂ in the reaction 2H₂ + O₂ → 2H₂O, what are the theoretical yield and the limiting reactant?
True [87]

Answer:

Now, we have to determine the limiting reagent.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g of

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.But 29 g is the given amount. Amount of oxygen unreacted = 29 - 24 = 5 g

7 0
3 years ago
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K Ho
Maru [420]

Answer : The mass of CS_2 is, 555.028 grams

Explanation :

First er have to calculate the concentration of S_2.

\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L

Now we have to calculate the concentration of CS_2.

The given balanced chemical reaction is,

                          S_2(g)+C(s)\rightleftharpoons CS_2(g)

Initial conc.         1.51       0         0

At eqm. conc.   (1.51-x)  (x)       (x)

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CS_2]}{[S_2]}

Now put all the given values in this expression, we get :

9.40=\frac{x}{(1.51-x)}

By solving the term 'x', we get :

x = 1.365 M

Concentration of CS_2 = x M = 1.365 M

Now we have to calculate the moles of CS_2.

\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole

Now we have to calculate the mass of CS_2.

Molar mass of CS_2 = 76 g/mole

\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g

Therefore, the mass of CS_2 is, 555.028 grams

3 0
4 years ago
If 3.00 g of limestone reacted, what mass of calcium chloride would be produced?
Levart [38]
CaCO₃ → CaCl₂
Ca = 40, C = 12, O = 16, Cl = 35.5
100 →111
3     →  x
x = 3.33g

8 0
3 years ago
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