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2 years ago

Given the balanced equation:

Chemistry

1 answer:

Svet_ta [14]2 years ago

8 0

Is there any answer choices before i get started on working out the problem

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What is the moisture content of hard gelatin capsule?

insens350 [35]

Answer:

The standard moisture content specification for hard gelatin capsules is between 13 % w/w and 16 % w/w.  This value can vary depending upon the conditions to which they are exposed: at low humidity's they will lose moisture and become brittle, and at high humidity's they will gain moisture and soften.

Explanation:

Hope this helps!

6 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Sauron [17]

Answer:

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the concentrations or changes in concentration in that stage. For this reaction:

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I 2.00 0 0

C -2x +x +x

E 2.00 - 2x x x

Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".

$Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852$

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0

3 years ago

A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance

sasho [114]

Answer:

$W\geq 2.1x10^{-19}J$

Explanation:

Due to Coulomb´s law electric force can be described by the formula $F=K\frac{q_{1}.q_{2}}{r^{2}}$ , where K is the Coulomb´s constant ( $9x10^{9} N\frac{m^{2} }{C^{2} }$ ), $q_{1}$ = Charge 1 (Na+ in this case), $q_{2}$ is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is $W=W_{f} -W_{i}$ .

so we have $W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]$

Given that ri= 1.1nm= $1.1x10^{-9}m$ and rf= infinite distance

$W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J$

6 0

3 years ago

F2 is light yellow/colorless in hydrocarbon solvent. I2 is pink. If a student combines fluorine water with NaI (sodium iodide) i

Tamiku [17]

F₂ + 2 NaI → 2 NaF + I₂
It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F₂ is present)

5 0

3 years ago

If Santa Claus is in the beach then what will you call him?

ahrayia [7]

Answer:

good question..... lemme think now LOL

6 0

3 years ago