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hoa [83]
3 years ago
15

What are the products of the reaction: Be(s)+ Cl2(g)⟶

Chemistry
1 answer:
Vikki [24]3 years ago
4 0

Answer:

BeCl2

Explanation:

Be+Cl2=BeCl2

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In negative adsorption the concentration of the solution is​
trapecia [35]

Answer:

This is called negative adsorption. Hence if the concentration of the adsorbate is less on the surface of the adsorbent than in the bulk, it is known as negative adsorption.

Explanation:

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What equation represents the pressure of a gas under ideal conditions?
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P=nRTV
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Need help Quickly Are volcanos and earthquakes both constructive and destructive forces
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yes

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3 years ago
2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo
pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

<u>Chemical reaction involved</u>: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K    (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol)     (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2

<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}

\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)

\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)

<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>

4 0
3 years ago
You need to make 10 mL of 2 mg/ml solution of protein and you have 25 mg/mL solution. How much protein solution and water do you
meriva

Answer:

0.8 mL of protein solution, 9.2 mL of water

Explanation:

The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:

C₁V₁ = C₂V₂

We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.

V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL

Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:

(10 mL - 0.8 mL) = 9.2 mL

6 0
3 years ago
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