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Sedbober [7]
2 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]2 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.

You might be interested in
2 attempts left select the single best answer. The radii of the lithium and magnesium ions are 76 pm and 72 pm, respectively. Wh
Lina20 [59]

The ionic character of any compound depend on the lattice energy as well as the electronegativity of element present in that compound.

More would be the lattice energy more would be ionic nature of that compound.

The lattice energy of any compound is inversely proportional to the ionic radii cation and anion.

In given case the ionic radii of oxide in both oxides would be equal therefore the lattice energy only depend on the ionic radii of cation.

Lattice energy (U)     =   \frac{1}{Ionic radii}

As the radii of Magnesium less then radii of lithium therefore lattice energy of Magnesium oxide would be more than lithium oxide.

Hence, MgO would be more ionic in nature than Li_{2}O

8 0
2 years ago
All of the following reactions can be described as displacement reactions except:____________.
Lady_Fox [76]

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>

Correct option = b.

8 0
2 years ago
After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?
FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: <em>Mg₃N₂</em>

<em></em>

I hope it helps!

5 0
1 year ago
At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same condition
yan [13]

Answer:

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H

-241.8kj -------eqn(1)

H2(g)+ 1/2 O2(g) ------>H2O(l).

Δ H =285.8kj ---------eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) ------>>H2O(g)---------------eqn(3)

But the equation from eqn(2) the eqn does go with vaporization so we can re- write as

H2O ------> H2(g)+ 1/2 O2(g)

Δ H= 285.8kj ---------------eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

5 0
2 years ago
Exactly 56 grams of iron is mixed with 156 grams of oxygen. The elements are heated and they react. What best describes which re
Mars2501 [29]

Answer:

Explanation:

The chemical expression for the reaction between iron and oxygen is:

4Fe(s)  +  3O₂ (g)  \to 2Fe₂O₃ (s)

The number of moles of Fe = mass of Fe/ molecular mass of Fe

The number of moles of Fe = 56 g/ 55.845 g/mol

The number of moles of Fe = 1.002 moles of Fe

The number of moles of oxygen = mass of oxygen/ molecular mass of oxygen

The number of moles of oxygen = 156 g /32 g/mol

The number of moles of oxygen = 4.875 moles of oxygen

Assume that Fe is the limiting reactant, the number of Fe₂O₃ can be calculated as:

moles of Fe₂O₃ = 1.002 mole of Fe × 2 moles of Fe₂O₃/ 4 moles of Fe

moles of Fe₂O₃ = 0.501 mole of Fe₂O₃

Assume that O₂ is the limiting factor, the number of Fe₂O₃ is:

moles of Fe₂O₃ = 4.875 moles of O₂ × 2 moles of Fe₂O₃/ 3 moles of O₂

moles of Fe₂O₃ = 3.25 mole of Fe₂O₃

Thus, after the reaction is complete, Fe and O₂ contain different moles of Fe₂O₃. Only Fe gets consumed in the reaction and it is the limiting factor.

8 0
1 year ago
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