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Tamiku [17]
1 year ago
15

A children's liquid medicine contains 100 mg of the active ingredient in 5 mL. If a child should receive 150 mg of the active in

gredient, how many milliliters of the medicine should the child be given? For the purposes of this question, assume that these numbers are exact. The package states that one teaspoon (tsp) is approximately equal to 5 mL. Calculate the number of teaspoons that the child should receive. Use the given approximation as your conversion factor.
Chemistry
1 answer:
likoan [24]1 year ago
4 0

Answer:

Explanation: A children's liquid medicine contains 100 mg of the active ingredient in 5 mL . If a child should receive 150 mg of the active ingredient, how many milliliters

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abruzzese [7]

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3 years ago
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Uranium, an important component of both nuclear weapons and nuclearreactors, has two major isotopes, U-238, which has a half-lif
elixir [45]

Answer:

Hello your question is poorly written below is the well written question

Uranium, an important component of both nuclear weapons and nuclear reactors, has two major isotopes, U-238, which has a half-life of  approximately 4.5 billion years, and U-235, which has a half-life of approximately 700 million years. Both were present in equal amounts at the time of the creation of the Earth, 4.5  billion years ago. How many years after the creation of the Earth had the amount of radiation from uranium decayed to half the amount present at the time of the creation of the Earth

Answer : 140 billion years

Explanation:

Given that :

U-238   h1/2 = 4.5 billion years

U-235   h1/2 = 700 million years

At the beginning both Isotopes where present in equal amount

Determine the T years before the amount of Uranium decays to Half

T =   ?    N'2 = N1 / 2

we know that N = No ( 1/2 )^h   where h = time / half-life time

attached below is the detailed solution of the given problem

6 0
3 years ago
Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
Marizza181 [45]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

The equation for the entropy change of the above reaction is:  

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]

We are given:

\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = \frac{173.008}{1}\times 2.25=432.52J/K

Hence, the value of \Delta S^o for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0
3 years ago
QUESTION 3 (a) Ammonium sulphate, (NH),50, is a soluble salt and it is used in agriculture as fertiliser. 5 g of ammonium sulpha
nataly862011 [7]

Answer:

The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)

The number of moles = 5 g / 132.14 g/mol = 0.038 mol

The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23

the number of positive ions present in the ammonium sulphate solution:

2 positive ions for every 1 molecule of (NH₄)₂SO₄

so 2 x 2.29x10^23 = 4.58x10^23

the number of negative ions present in the ammonium sulphate solution

1 negative ion for every 1 molecule of (NH₄)₂SO₄

so 1 x 2.29x10^23 = 2.29x10^23

the total number of ions present in the ammonium sulphate solution​

4.58x10^23 + 2.29x10^23 = 6.87x10^23

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3 years ago
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Answer:

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