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AleksandrR [38]
3 years ago
14

At a dosage of 20 mg per kg of body weight, how many mg of that drug should a 175 lb adult receive

Chemistry
1 answer:
zubka84 [21]3 years ago
5 0

It all depends on what type of drug.


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How many grams of CrSO3 are in 2.4 moles of CrSO3
castortr0y [4]

Answer: 316.8 g CrSO3

Explanation: Solution:

2.4 moles CrSO3 x 132 g CrSO3 / 1 mole CrSO3 = 316.8 g CrSO4

The conversion factor is 1 mole of CrSO4 is equal to its molar mass which is 132 g CrSO3

4 0
3 years ago
Which of the following statements concerning the measured number 0.3030
Dennis_Churaev [7]

Answer:

C

Explanation:

If you move the decimal to the left it transforms the expression into scientific notation.

8 0
3 years ago
Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden
LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

5 0
3 years ago
In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0
3 years ago
Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. use th
Nataly [62]

There are two naturally occurring isotopes of gallium:  mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:

%Ga-71=100-60.11=39.89%

Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.

Thus, in this case:

Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)

From the periodic table, atomic mass of Ga is 69.723 amu.

Putting the values,

69.723 amu=(68.9256 amu)(\frac{60.11}{100})+X(\frac{39.89}{100})

Thus,

69.723 amu=41.4312 amu+X(\frac{39.89}{100})

Rearranging,

X=\frac{69.723 amu-41.4312 amu}{0.3989}=70.9246 amu

Therefore, mass of Ga-71 isotope is 70.9246 amu.

7 0
3 years ago
Read 2 more answers
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