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Lisa [10]
2 years ago
13

FLAG A nurse is preparing to administer morphine 0.05 mg/kg intermittent IV bolus to a newborn who weighs 3 kg. Available is mor

phine 0.5 mg/mL injection. How many mL should the nurse administer? (Round the answer to the nearest tenth. Use a leading zero if it applies. Do not use a trailing zero.) ​
Chemistry
1 answer:
erica [24]2 years ago
6 0

The quantity of morphine, the nurse should administer is 0.3 ml.

<h3>What is morphine?</h3>

Morphine is a drug which comes under the pain relieving drugs.

The quantity of morphine is 0.05 mg/kg

The weight of the baby is 3 kg.

Available morphine is 0.5 mg/mL.

0.05 x 3 = 0.15 mg

then, 0.15 mg / 0.5 = 0.3

Thus, the quantity of morphine, the nurse should administer is 0.3 ml.

Learn more about morphine

brainly.com/question/10665765

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A boy found a solid metal box in his backyard. The box had been buried for so long, it was difficult to determine from what the
Alex777 [14]

Answer:

Box is made up of <em>copper</em>, because density is <em>8.96  g/cm³.</em>

Explanation:

Given data:

Volume of box = 17.63 cm³

Mass of box = 158 g

Which metal box is this = ?

Solution:

First we will calculate the density of box then we will compare it with the density value of given metals.

d = m/v

d = 158 g/ 17.63 cm³

d = 8.96  g/cm³

The calculated density is similar to the given density value of copper thus box is made up of copper.

4 0
4 years ago
Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon
natali 33 [55]

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= \frac{100.0 \ g}{114.33\  g/mole}

= 0.8747 \ moles

We know that 16 moles of CO_2 formation associates with -11018 kJ of heat, then

0.8747 moles CO_2 formation associates with,

= -\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat

= -0.0547\times 11018

= -602.69 \ KJ \ heat

8 0
3 years ago
A sample of pure water has a hydronium concentration of 1.0 × 10-7 M. What is the pH of the water?
stepan [7]
PH of the water would be 6.79
5 0
3 years ago
Read 2 more answers
2) What is the concentration of NaCl in a solution prepared by diluting 56.98 ml of 0.5894 M stock
RSB [31]

Answer:

The NaCl concentration will be 0.03 M.

Explanation:

Given data:

Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)

Initial concentration = M₁= 0.5894 M

Final volume = V₂= 1.20 L

Final concentration = M₂= ?

Solution:

By diluting the solution volume of solution will increase while number of moles of solute remain the same.

Formula:

Initial concentration × Initial volume  = Final concentration × Final volume

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 0.5894 M × 0.05698 L / 1.20 L

M₂ = 0.0336 M /1.20  

M₂ = 0.03 M

3 0
4 years ago
What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) =&gt; 2
zavuch27 [327]

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, \frac{6}{635}\times 44.7=0.423mol of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

7 0
4 years ago
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