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Sergeeva-Olga [200]
2 years ago
13

Children’s chewable tylenol contains 80 mg of acetaminophen per tablet. if the recommended dosage is 10 mg/kg, how many tablets

are needed for a 53−lb child?
Chemistry
2 answers:
vovikov84 [41]2 years ago
8 0

Answer:

3 tablets are needed for a 53-lb child

Explanation:

1 lb = 0.4536 kg

So, 53 lb = (53\times 0.4536)kg=24 kg

So, weight of the child is 24 kg

For 1 kg body, recommended dosage is 10 mg

Hence, for 24 kg body, recommended dosage is (10\times 24)mg or 240 mg

80 mg of acetaminophen is present in 1 tablet

So, 240 mg acetaminophen is present in \frac{1}{80}\times 240 tablets or 3 tablets.

So, 3 tablets are needed for a 53-lb child

eduard2 years ago
6 0

53 pounds divided by 2.2 kilograms to get the converted weight which is 24.09 kilograms. Round this off to 24 kgs. 10 mg/kg multiply this to 24 kg, it would be 240 mg. 240 mg divided by 80 is 3. Therefore, a 53−lb child requires to have a dosage of 3 tablets.

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Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) form a self-igniting mixture that has been used as a rocket propellant. The rea
Alexxx [7]

Answer:

Explanation:

An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.

Chemical equation:

1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O

2) Hydrazine ( N₂H₄)  is being oxidized

Dinitrogen tetroxide N₂O₄ is being reduced

3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)

5 0
2 years ago
What is the maximum number of hydrogen atoms that can be covalently bonded in a molecule containing two carbon atoms?a) 2b) 3c)
aksik [14]

Answer:

           The correct answer is Option-D (6 Hydrogen atoms).

Explanation:

                  Carbon atom has a unique property of linking to it self and making a chain of carbon called as Catenation. In a molecule having two carbon atoms there must be a bond between the two carbon atoms. The bond can be saturated (single bond) or unsaturated (double or triple bond). But as the statement states maximum number of hydrogen atoms so, we will asume the bond between two carbon atoms to be single.

                  As carbon has four valence electrons so, it has the ability to make four single bonds. In given molecule each carbon has already made a single bond with another carbon therefore, each carbon is left with three more unpaired electrons which will make covalent bond with three hydrogen atoms each as shown in attached structure.

8 0
2 years ago
Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p
Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

  • 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
  • The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}

8 0
2 years ago
The number of p atoms in 1.0 g of ba3(po4)2 is:
marta [7]
First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002

Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³

2.41 x 10²¹ atoms of phosphorus are present
3 0
2 years ago
If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln
Ksju [112]
Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
5 0
2 years ago
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