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3 years ago

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From the time a drug is discovered, a drug is given this name which is assigned by the International Union of Pure and Applied C

hemistry (IUPAC).
A. Chemical name
B. Brand name
C. Trade name
D. Generic name

1 answer:

sweet-ann [11.9K]3 years ago

7 0

Answer:

A. Chemical name

Explanation:

The Chemical Name is the name generated by applying the naming conventions of the International Union of Pure and Applied Chemistry (IUPAC). The correct names of compounds contained in drug products can be formed by implementing these rules

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Which of the atoms shown has an atomic number of 4?

spin [16.1K]

Answer:

B. Be

Explanation:

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3 years ago

How many liters of carbon dioxide will 0.5 moles of lithium hydroxide (LiOH) absorb?

Elden [556K]

Answer : The volume of carbon dioxide will be, 5.6122 L

Solution : Given,

Moles of LiOH = 0.5 moles

Molar mass of carbon dioxide = 44 g/mole

Density of carbon dioxide = 0.00196 g/ml

First we have to calculate the mass of carbon dioxide.

The balanced reaction will be,

$2LiOH+CO_2\rightarrow Li_2CO_3+H_2O$

From the reaction we conclude that

As, 2 moles of LiOH absorbs 44 grams of carbon dioxide

So, 0.5 moles of LiOH absorbs $\frac{0.5moles}{2moles}\times 44g=11$ grams of carbon dioxide

Mass of carbon dioxide = 11 g

Density of carbon dioxide = 0.00196 g/ml

Now we have to calculate the volume of carbon dioxide.

$Density=\frac{Mass}{volume}$

$0.00196g/ml=\frac{11g}{volume}$

Volume of carbon dioxide = 5612.24 ml = 5.6122 L (1 L = 1000 ml)

Therefore, the volume of carbon dioxide will be, 5.6122 L

5 0

3 years ago

Which statements describe how heat flows in foil?

LenKa [72]

Answer: c

Explanation:

8 0

3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

7 0

3 years ago