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Anettt [7]
3 years ago
5

1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-l

b woman. Show all your calculations. (1lb = 453.6 g)
2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
7 0

Answer:

1. 0.138g of valium would be lethel in the woman

2. 125mg/min is the drip of the patient

Explanation:

1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.

A person that weighs 200lb requires:

200<u>lb</u> × (453.6<u>g</u> / <u>1lb</u>) × (1kg / 1000<u>g</u>) = <em>90.72kg (Weight of the woman in kg)</em>

90.72kg × (1.52mg / kg) =

137.9mg ≡

<h3>0.138g of valium would be lethel in the woman</h3>

2. The IV contains 1.5g = 1500mg/mL.

If the patient is receiving 5.0mL/h, its rate in mg/h is:

5.0<u>mL</u>/h × (1500mg/<u>mL</u>) = 7500mg/h

Now as 1h = 60min:

7500mg/<u>h</u> × (1<u>h</u> / 60min) =

<h3>125mg/min is the drip of the patient</h3>
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Draw the conjugate base for the Brønsted-Lowry acid-base reaction that occurs when the following acid reacts with water. Show al
DanielleElmas [232]

Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

FC = X - (Y + Z/2)

Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.

7 0
3 years ago
Read 2 more answers
Help me, I’m failing chem
ladessa [460]

Answer:

b

Explanation:

[H3O+] = 10-pH = 10-3.4 ≅ 3.981 x 10^-4 moles/liter

5 0
3 years ago
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I really need help with this question please help me.
oksano4ka [1.4K]
I would say A because it’s the only one that doesn’t conduct electricity. Ionic compounds conduct electricity, and covalent compounds do not.
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