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Rashid [163]
3 years ago
14

The top of the syringe is a circle. You need to

Chemistry
1 answer:
AfilCa [17]3 years ago
4 0

Answer:10.2 cm2

Explanation:

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fredd [130]

Answer:

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Explanation:

6 0
2 years ago
Read 2 more answers
I need help on thiss
Sergio039 [100]

The reaction is not balanced

<h3>Further explanation</h3>

Given

Reaction

2Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

Required

The number of atoms

Solution

In a balanced chemical equation, the number of atoms in the compound that reacts (the reactants and products) will have the same number

Reactants : Fe(s)+O₂(g)

Fe = 2 atoms

O = 3 x 2 = 6 atoms

Products : Fe₂O₃(s)

Fe = 2 x 2 = 4 atoms

O = 2 x 3 = 6 atoms

The reaction is not balanced because the number of Fe atoms is not the same

The balanced reaction should be:

4Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

8 0
3 years ago
Solve for Va<br><br>MaVa=MbVb​
PolarNik [594]

Answer:

Va = (MbVb)/Ma

Explanation:

Divide both sides by Ma and voila!

8 0
3 years ago
Which of the following is true?
Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

ΔG = ΔH - TΔS

  • When ΔG < 0 the reaction is spontaneous in the forward direction.
  • When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

  • When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS  term is greater in magnitude than ΔH. This condition is met when T is large.

3 0
3 years ago
If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

5 0
3 years ago
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