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torisob [31]
3 years ago
15

Chemicals A and B react according to the equation A + B → C. How will the concentrations of each component of the reaction chang

e over time?
The concentrations of A and C will increase, while B will decrease.

The concentrations of A and B will decrease, while C will increase.

The concentrations of A, B, and C will decrease.

The concentrations of A, B, and C will increase.
Chemistry
2 answers:
Firlakuza [10]3 years ago
7 0
The concentration of a and b will decrease while c will increase provided all other physical quantities are kept constant in the reaction
Irina18 [472]3 years ago
3 0

Answer:

The correct answer is ;The concentrations of A and B will decrease, while C will increase.

Explanation:

A + B → C

In the given reaction reactant A and B reacts with each other to form product C. As the reaction proceeds the concentration of A and B starts decreasing and concentration of product C increases.

In the end of the reaction concentration of C becomes greater than that of the  A and B concentration

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Formation of water: 2H2 + 1 O2 --> 2H2O
noname [10]

Answer:

2.2 moles H2O

Explanation:

35g O_2 \mbox{ \cdot }\frac{1mol}{32g/mol} \mbox{ \cdot }\frac{2mol H_2O}{1mol O_2}= 2.1875, which rounds to about 2.2

7 0
2 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
How do I solve this problem?
JulijaS [17]
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
5 0
3 years ago
Which term best describes a unit of carbon dioxide?
Lelechka [254]

Answer:

Molecule.

Explanation:

carbon and fluorine. Based on the bonding, a unit of carbon dioxideis described as a molecule.

7 0
2 years ago
Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0L. What temperature must be maintained
Morgarella [4.7K]
Given:

P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?

Assume that the gas ideally, thus, we can use the ideal gas equation:

PV = nRT

where R = 0.0821 L atm/mol K

123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T

solve for T 

T = 295.72 K<span />
5 0
3 years ago
Read 2 more answers
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