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aev [14]
4 years ago
12

A dose of aspirin of 5.0 mg per kilogram of body weight has been prescribed to reduce the fever of an infant weighing 8.5 pounds

. The number of milligrams of aspirin that should be administered
Chemistry
2 answers:
Tanya [424]4 years ago
5 0
<h3><u>Answer</u>;</h3>

19 mg

<h3><u>Explanation</u>;</h3>

dose = 5.0 mg/kg  

weight of infant = 8.5 pounds

Convert 8.5 pounds to kg: 8.5 pounds = 3.86 kg

Therefore;

The dose to give the infant is:  

= (5.0 mg/kg) x (3.86 kg)

= 19.3 mg  

<u>≈ 19 mg</u>

dolphi86 [110]4 years ago
5 0

Answer:

19.2737 milligrams of aspirin that should be administered.

Explanation:

Mass of an infant = 8.5 pounds = 8.5 × 0.4535 kg=3.8547 kg

Aspirin dosage per 1 kilogram = 5 mg

If an infant weights 3.8547 kg, the dosage required will be;

3.8547 × 5 mg=19.2737 mg[/tex]

19.2737 milligrams of aspirin that should be administered.

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He standard reduction potentials of lithium metal and chlorine gas are as follows: reaction reduction potential (v) li+(aq)+e−→l
AURORKA [14]

 2 Li(s)  +Cl₂→  2 Li⁺ (aq)  + 2Cl⁻ (aq)

The cell potential   of  the reaction above   is +4.40V

<em><u>calculation</u></em>

Cell  potential  =∈° red - ∈° oxidation

in  reaction above  Li  is  oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04

  Cl  is  reduce   from oxidation  state 0 to -1 therefore  the ∈°red = +1.36 V

cell  potential is therefore = +1.36 v -- 3.04  = + 4.40 V

3 0
3 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
What is the average of atomic mass of carbon if 98.90% of the atoms are c-12 and 1.100% are c-13 atoms?
Paladinen [302]

Answer:

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5 0
2 years ago
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g)   (1)

The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

5 0
3 years ago
I don’t get it plz help
d1i1m1o1n [39]
D=m/v. (Density equals mass over volume).
So in your case density = 4.2/6
Which equals 0.7
Your units will be grams over milliliters

So your answer is 0.7 g/ml
3 0
3 years ago
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