(2C+2+N-X-H)/2
(2*7+2+0-0-8)/2
4
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>
Answer:
1.09 grams
Explanation:
According to the following chemical equation:
HF + NaNO₃ -> HNO₃ + NaF
1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:
MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF
1 mol HF x 19.9 g/mol HF = 20 g
MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF
1 mol NaF x 42 g/mol NaF = 42 g
Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:
20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF
Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.
