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Sedbober [7]
1 year ago
12

29. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servi

ngs of this cereal must be eaten to consume 0.0278 moles of sugar?
Chemistry
1 answer:
adoni [48]1 year ago
3 0

Answer:

The answer is: 51.8 g (86% of serving size)

Explanation:

In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:

11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol

We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):

0.032 mol ------------ 60 g serving

0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol

                                x= 51.8 g

So,  lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).

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5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperatu
kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
1 year ago
30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is
grigory [225]

Answer:

             Mass  =  6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

                                        Molarity  =  Moles ÷ Volume

Solving for Moles,

                                        Moles  =  Molarity × Volume

Putting Values,

                                        Moles  =  0.05 mol.L⁻¹ × 2.0 L

                                        Moles  =  0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

                                        Moles  =  Mass ÷ M.mass

Solving for Mass,

                                        Mass  =  Moles × M.mass

Putting values,

                                        Mass  =  0.1 mol × 61.83 g.mol⁻¹

                                        Mass  =  6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.

7 0
2 years ago
A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr
s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

P=p_{N_2}+p_{O_2}+p_{Ar}

p_{N_2}=P\times \chi_{N_2}

p_{O_2}=P\times \chi_{O_2}

p_{Ar}=P\times \chi_{Ar}

where,

P = total pressure = 3.9 bar

p_{N_2} = partial pressure of nitrogen gas  

p_{O_2} = partial pressure of oxygen gas  

p_{Ar} = partial pressure of argon gases  

\chi_{N_2} = Mole fraction of nitrogen gas  = 0.25

\chi_{O_2} = Mole fraction of oxygen gas  = 0.65

\chi_{Ar} = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar

Partial pressure of oxygen gas :

p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar

Partial pressure of argon gas :

p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar

7 0
2 years ago
This same chemistry student has a weight of 155 lbs. What is the student’s weight in
Virty [35]

Answer:

the same chemistry student has a weight of 155lbs what is the student weight in grams? (16 oz= 1 lb, 1 oz= 28.34g)

Explanation:1 lb = 16oz, so multiply your pounds by 16 to get you ounces of the student, then multiply by 28.34 to get grams

155 X 16 X 28.34 = 70283.2

8 0
1 year ago
Read 2 more answers
A student was asked to find the density of an unknown object. The student found that the mass of the object was 35.49 g. The stu
marusya05 [52]

Answer:

4.29 /mL

Explanation:

density= mass/volume

8 0
1 year ago
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