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sweet [91]
3 years ago
7

You are to give ampicillin, with a recommended dose of 25 mg/kgmg/kg , to a child with a mass of 24 kgkg . If stock on hand is 2

50 mg/capsulemg/capsule, how many capsules should be given?
Chemistry
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

2.4 capsules

Explanation:

You are to give ampicillin, with a recommended dose of 25 mg/kg (25 mg of ampicillin per kg of body weight), to a child with a mass of 24 kg. The mass of ampicillin to be given is:

24 kg body mass × (25 mg ampicillin/ 1 kg body mass) = 600 mg ampicillin

The stock on hand is 250 mg/capsule (250 mg ampicillin per capsule). The number of capsules required to supply 600 mg of ampicillin is:

600 mg ampicillin × (1 capsule/250 mg) = 2.4 capsule

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Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-
Semenov [28]

Answer:

d) PO4^3-, HPO4^2-

Explanation:

Basically, an acid and a base which differs only by the presence or absence of  proton (hydrogen ion) are called a conjugate acid-base pair.

a) HI, I

This is incorrect. For the acid, HI the conjugate base is I⁻ ion.

b) HCHO2, SO4^2-

This is incorrect, there's no relationship between both entities.

c) CO3^2-, HCI

This is incorrect, there's no relationship between both entities.

d) PO4^3-, HPO4^2-

This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair

6 0
3 years ago
Unit: Stoichiometry
Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0
3 years ago
What os the formula for hexaboron tetrabromide
LekaFEV [45]

Answer:

B6Br4

Explanation:

Chemical formula gives a representation of the number of atoms that each element in a compound or molecule. In this question which is asking to give the formula of hexaboron tetrabromide, the hexa means 6 while the tetra means four.

Since the number of atoms of each element is written as a prefix before the main element, this means that the chemical formula of hexaboron tetrabromide will be B6Br4.

6 0
3 years ago
La is element 57 on the periodic table a sample contains 2.82 * 10€25 power atoms of La calculate the amount of LA
vodka [1.7K]

Answer:

n=46.8molLa\\\\m=6.50x10^3gLa

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate both moles and grams of lanthanum by using the Avogadro's number as a relationship of atoms to moles and its atomic mass as a relationship to moles to grams to obtain the following:

n=2.82x10^{25}atomsLa*\frac{1molLa}{6.022x10^{23}atomsLa}=46.8molLa\\\\m=46.8molLa*\frac{138.9gLa}{1molLa}  =6.50x10^3gLa

Regards!

5 0
3 years ago
Consider the following unbalanced reaction: P4(s) + F2(g) → PF3(g) What mass of fluorine gas is needed to produce 120. g of PF3
Studentka2010 [4]

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

P_{4}+6F_{2}-->4PF_{3}

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =\frac{mass}{molarmass}=\frac{120}{88}=1.364mol

The moles of fluorine gas required = \frac{4X1.364}{6}=0.91

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = \frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g

4 0
3 years ago
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