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Sedbober [7]
3 years ago
6

a nurse practitioner orders Medrol to be given 1.5 mg/kg of body weight. if a child weighs 72.6lb and the available stock of Med

rol is 20.mg/mL, how many mL do you give to the child
Chemistry
1 answer:
Harrizon [31]3 years ago
8 0
Answer is: 2,469 mL give to the child.
The mass m in kilograms (kg) is equal to the mass m in pounds (lb) times 0,45359237: m(child) = 72,6 · 0,045359237 = 32,93 kg.
m(Medrol) = 32,93 kg · 1,5 mg/kg.
m(Medrol) = 49,39 mg.
d(Medrol) = 20,0 mg/mL.
V(Medrol) = m(Medrol) ÷ d(Medrol).
V(Medrol) = 49,39 mg ÷ 20 mg/mL.
V(Medrol) = 2,469 mL.

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Plsss help this midterm ends 12.30
Alona [7]

Answer:

We know that,

Density= mass upon volume.

mass= 21.93g

Volume= 49.3 - 46.3

= 3 cm³

Density = 21.93÷ 3

= 7.31 g/cm³

Hope it helps :))

5 0
3 years ago
It
ivann1987 [24]

Answer:

His mistake was that he also gave the plants  in the music room some fertilizer

Explanation:

When carrying out a reliable experiment, the only variable should be the question being tested. The question here is whether playing music for plants helps them grow faster. Therefore, all the plants should be grown in exactly the same conditions apart from the presence or absence of music.

However, Jo also added fertilizer to the plants in the room where music was playing. Therefore, we don't know whether the fertilizer or the music caused the plants to grow faster.

Instead, Jo should have not given fertilizer to any of the plants, or given fertilizer to all the plants.

5 0
3 years ago
How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.
Slav-nsk [51]
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0
3 years ago
Can anyone please help
Naddik [55]
All of them are reactants
7 0
3 years ago
The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o
telo118 [61]
If you clear volume in the density equation:

\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3

V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3
6 0
3 years ago
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