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1 year ago

How many molecules are there in 9.34 grams of LiCL

1 answer:

7 0

Assuming the question refers to LiCl (Lithium chloride) which has a molecular weight 42.39. Avogadro's constant states there are 6.022 141 79x1023 molecules per mole 9.34 g LiCl is 9.34/42.39 mole (0.220 mole) LiCl 
The number of molecules is therefore 6.022 141 79x1023x 0.220 =1.326x1023 molecules

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How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to

saw5 [17]

Answer:

Drug calculation

If we have 45g of clobetasol = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

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1 year ago

Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to

Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode $(m_{ch})$ is given as :

$\frac{m_{ch}}{M} ZF = It$

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; $(m_{ch})$ = $(Ad) \rho$ ; replacing that into above equation; we have:

$\frac{(Ad) \rho}{M} ZF = It$ ---- equation (1)

where;

A = area

d = thickness

$\rho$ = density

From the above equation (1); The time required for coating can be calculated as;

$[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t$

$t = \frac{2100}{20}$

= 105 s

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2 years ago

The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

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1 year ago

Which traits does the common garter snake have that might be adaptive for the environment where it lives?

Licemer1 [7]

Answer:

As the garter snake can be found almost in any kind of habitat, what makes them be able to survive in any environment include:

1. They hibernate to increase their chances of survival in unfavorable weather conditions.

2. They can blend with the background of any environment especially grass to escape being eaten.

3. They produce an odor that is usually unpleasant especially when about to be attacked.

Explanation:

The garter snakes are distinguished by the three stripes running the length of their body and can often be found in forests, places that are even close to water bodies, and almost any place, even in holes.

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1 year ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$