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dedylja [7]
4 years ago
11

A doctor has ordered that a patient be given 15g of glucose, which is available in a concentration of 75.00g glucose/1000.0 mL o

f solution.What volume of solution should be given to the patient?
Chemistry
2 answers:
AnnZ [28]4 years ago
8 0
Given the concentration (C) in mass per volume, the volume (V) of the solution is obtained by dividing the mass (m) by this concentration (C). So, from the given,
                         V = C / m   = (15 g) / (75 g / 1000 mL)
The numerical value of V is equal to 200 mL. 
Lynna [10]4 years ago
7 0
I would start by converting g/mL to g/L
75.00g/1000.0mL = 75.00g/L
Stoichiometrically, flip 75 onto the bottom so that grams cancel out and we are left with the number of L required. 
(L/75.00g)(15g) -> this is essentially dividing 15g by 75g, which cancels the unit g, leaving us with 0.2L. If the question requires an answer in mL, just multiple the number of L by 1000. 
The patient requires 200mL of glucose solution to receive his 15g of glucose. 
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The volume of a gas is 1.15 L at STP. At what temperature would the volume of the gas increase to 1.56 L, if the pressure is dec
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A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

<h3>What is an ideal gas?</h3>

An ideal gas is a gas whose behavior can be explained through ideal gas laws. One of them is the combined gas law.

A gas occupies 1.15 L (V₁) at STP (T₁ = 273,15 K and P₁ = 760 mmHg). We can calculate the temperature (T₂) at which V₂ = 1.56 L and P₂ = 650 mmHg, using the combined gas law.

\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\\T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1} = \frac{650mmHg \times 1.56 L \times 273.15K}{760mmHg \times 1.15 L} = 317 K

A gas occupies 1.15 L at standard pressure and temperature and 1.56 L at 317 K and 650 mmHg, assuming ideal behavior.

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2 years ago
if you are using 657.0 grams of oxygen gas to react with calcium how many gram of calcium oxide will u product
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The amount in grams of CaO that will be produced is 2,302.67824 grams

Explanation:

The reaction between calcium and oxygen to produce calcium oxide (quicklime) can be presented as follows;

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2 moles of calcium, Ca, react with 1 mole of oxygen gas, O₂, to produce 2 moles of CaO

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The number of moles, 'n', of O₂ in 657.0 grams of oxygen is therefore;

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1 mole of O₂ produces 2 moles of CaO, therefore, 20.53125 moles of oxygen will produce 2 × 20.53125 or 41.0625 moles of CaO

The molar mass of CaO = 56.0774 g/mol

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Mass of substance = (The number of moles of a substance) × (Molar mass of substance)

The mass of CaO produced = 41.0625 moles × 56.0774 g/mol = 2,302.67824 g.

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