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nadya68 [22]
2 years ago
14

An order calls for Hytone 1% cream, Lotrimin 1% cream, aa q.s. How many grams of Lotrimin cream would be needed to make a total

of 60 g of cream?
Chemistry
2 answers:
Ainat [17]2 years ago
7 0

The words aa q.s. is an abbreviation which is used in pharmacy. This is usually written by prescribers and then translated by pharmacists. The aa is derived from Latin which means “of each” while q.s. is also derived from Latin which means “a sufficient quantity”. This simply means that there is equal amount of Hytone 1% cream and Lotrimin 1% cream in the total cream formulation.

Since we are to make a total of 60 g of cream, therefore 50% of it would be Lotrimin cream.

Amount of Lotrimin cream needed = 0.5 (60 g)

<span>Amount of Lotrimin cream needed = 30 grams</span>

Novay_Z [31]2 years ago
7 0

Answer:

0.6~g~of~lotrimin

Explanation:

We have to keep in mind that the amount of lotrimin in the cream is 1%. Now the total of cream we want to make is 60 g. So, we have to ask us, what is 1% of 60 g? If we keep in mind that if we have 1% indicates that in 100 g of cream we have 1 g of lotrimin we can do the calculation:

60~g~of~cream\frac{1~g~of~lotrimin}{100~g~of~cream}

0.6~g~of~lotrimin

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A 12.2-g sample of x reacts with a sample of y to form 78.9 g of xy. what is the mass of y that reacted?
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We will assume that the only reactants are x and y and that the only product is xy.

Based on the law of mass conservation, mass is an isolated system that can neither be created nor destroyed.

Applying this concept to the chemical reaction, we will find that the total mass of the reactants must be equal to the total mass of the products,
therefore:
mass of x + mass of y = mass of xy
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2 years ago
In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq)
kenny6666 [7]

Answer:

The anode half reaction is : Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}

Reduction (cathode): Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)

------------------------------------------------------------------------------------------------------------

overall: Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)

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8 0
2 years ago
A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

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Read 2 more answers
A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,
xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

  • PT= 987 torr
  • PN₂= 44 torr
  • PAr= 486 torr
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Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

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