Convert the child weight (37.3 pounds) to kilograms
37.3 lb x 0.453 kg /1lb = "A kg"
multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him
A kg * 9.00 mg/1kg = "B mg"
calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL
B mg * 1mL/ 60.0 mg = C mL <span>oxcarbazepine</span>
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
"c. Number the chain of carbons, starting with the end nearest the multiple bond."
This choice means that the carbon atoms determine the name of the hydrocarbonsThank you for your question. Please don't hesitate to ask in Brainly your queries.
Scandium? Is that what you mean?
CH3COOH
Explanation:
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