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Angelina_Jolie [31]
3 years ago
13

A sample of sugar (C12H22O11) contains 1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample?

Chemistry
2 answers:
Tomtit [17]3 years ago
7 0
1 mole ------------ 6.02x10²³ molecules
? moles ----------- 1.505x10²³ molecules

(1.505x10²³) x 1 /  ( 6.02  x10²³ ) =

(1.505x10²³) / ( 6.02x10²³ ) =

= 0.25 mol

Answer A

hope this helps!
saveliy_v [14]3 years ago
6 0

Answer: Option (a) is correct.

Explanation:

Molecules of sugar given = 1.505 \times 10^{23}

According to Avogadro's number it is known that 1 mole contains 6.023 \times 10^{23}.

Therefore, calculate the moles of sugar present in the sample as follows.

   Number of moles =  \frac {1.505 \times 10^{23}}{6.023 \times 10^{23}}

           = 0.25 mol

Thus, the moles of sugar present in the sample is 0.25 mol.

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A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder
IRISSAK [1]

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A. Up

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C. Out

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a. The reaction in an exothermic reaction so this means heat is given off. If the cylinder is thin enough heat will transfer to the water bath

b. Since the products will create heat which will increase pressure, the piston in an attempt to maintaining a constant pressure will move up to accommodate building pressure.

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7 0
4 years ago
Calculate the volume of chlorine at stp that would be required to act completely with 3.70g of dry slaked line
myrzilka [38]

Answer:

The volume required is 2.24L

Explanation:

The slaked lime Ca(OH)2, reacts with chlorine, Cl2, as follows:

6Cl₂(g) + 3Ca(OH)₂(s) → Ca(ClO₃)₂ (aq) + 2CaCl₂ (aq) + 6HCl (aq)

<em>Where 6 moles of chlorine react with 3 moles of slaked llime,</em>

<em />

To solve this question we must find the moles of slaked lime added. With these moles we can find the moles of chlorine required and its volume at STP as follows:

<em>Moles Ca(OH)2 - Molar mass: 74.093g/mol-</em>

3.70g * (1mol / 74.093g) = 0.0500 moles Ca(OH)2

<em>Moles Cl₂:</em>

0.0500 moles Ca(OH)2 * (6 mol Cl₂ / 3 mol Ca(OH)2) =

0.100 moles Cl₂

Now using PV = nRT

nRT / P = V

<em>Where n are moles: 0.100 moles</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273K at STP</em>

<em>P is pressure = 1atm at STP</em>

<em>And V is volume, our incognite:</em>

<em />

0.100mol*0.082atmL/molK*273K / 1atm = V

2.24L = Volume of Cl₂

<h3>The volume required is 2.24L</h3>
8 0
3 years ago
Perform the following for Part C of this lab:
kaheart [24]

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

<em>Titrant (NaOH) concentration: 1.0 M</em>

<em>Vinegar volume: 2.0 mL</em>

<em>Initial buret reading (initial NaOH volume): 0.1 mL</em>

<em>Final buret reading (final NaOH volume): 11.1 mL</em>

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mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%

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