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2 years ago

50 POINTS

Chemistry

2 answers:

NeX [460]2 years ago

6 0

Answer: C

Silver is an Excellent conductor

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lesya692 [45]2 years ago

3 0

Explanation:

c Silver is an excellent conductor

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18 ) Liquid A and Liquid B are clear liquids. They are placed in open containers and allowed to evaporate. When evaporation is c

arsen [322]

Answer:

Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture

Explanation:

5 0

3 years ago

Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is

Elis [28]

Answer:

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products =

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

We have :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

$v=V_{max}\times \frac{[S]}{K_m+[S]}$

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

8 0

3 years ago

Consider the reaction: I2(g)+Cl2(g)⇌2ICl(g) Kp= 81.9 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each condition: -

scoray [572]

Answer:

Part 1: - 1.091 x 10⁴ J/mol.

Part 2: - 1.137 x 10⁴ J/mol.

Explanation:

Part 1: At standard conditions:

At standard conditions Kp= 81.9.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.

Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.

For the reaction:

I₂(g) + Cl₂(g) ⇌ 2ICl(g).

Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.

∵ ΔGrxn = -RTlnKp

∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.

6 0

3 years ago

Read 2 more answers

The elements in groups 1a through 7a are

REY [17]

The elements in<span> groups 1a through 7a are called d. representative elements. The periodic table is organized according to increasing order of the atomic number of the elements. Groups have similar properties while the period of the elements determines the valence electrons in the outer shells.</span>

3 0

3 years ago

Read 2 more answers

The ratio of nitrogen to oxygen by mass in NO is 7.0:8.0. Identify the ratio of nitrogen to oxygen by mass in NO2 and N2O7 .

Rus_ich [418]

This problem is providing the ratio of nitrogen to oxygen by mass in nitrogen monoxide, NO, as 7.0:8.0 and asks for the same ratio but in NO₂ and N₂O₇. After doing the calculations, the results are 7.0:16.0 and 1.0:4.0 respectively.

<h3>Mass ratios:</h3>

In chemistry, one can calculate the mass ratios in chemical formulas according to the atomic mass of each atom. In such a way, one knows the mass ratio of nitrogen to oxygen in NO is 7.0:8.0 because we divide the atomic mass of nitrogen by oxygens:

$\frac{14}{16}=\frac{7.0}{8.0}$

Now, for chemical formulas with subscripts, one must multiply the atomic mass of the element by the subscript in the formula, which is the case of NO₂ and N₂O₇ as shown below:

$NO_2:\frac{14}{16*2}=\frac{14}{32} =\frac{7.0}{16.0} \\
\\
N_2O_7:\frac{14*2}{16*7}=\frac{28}{112} =\frac{1.0}{4.0}$

Therefore, the results for NO₂ and N₂O₇ are 7.0:16.0 and 1.0:4.0 respectively

Learn more about atomic masses: brainly.com/question/5566317

3 0

2 years ago