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gtnhenbr [62]
3 years ago
12

In laboratory experiment, a NOVDEC Student was

Chemistry
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

<em>The volume of the glucose solution to be prepared</em> = 500 cm^3

<em>Molarity of the glucose solution to be prepared</em> = 1 M

i. Molar mass of glucose (C_1_2H_6O_6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii.<em> mole = molarity x volume</em>. Hence;

amount (in moles) of the glucose solution to be prepared

                 = 1 x 500/1000 = 0.5 mole

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What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

8 0
3 years ago
What temperature air has low pressure?
finlep [7]

Explanation:

A low-pressure area, or "low", is a region where the atmospheric pressure at sea level is below that of surrounding locations. Low-pressure systems form under areas of wind divergence that occur in upper levels of the troposphere.

5 0
3 years ago
How many grams of carbon are present in 45.0 g of CCl4?
lisabon 2012 [21]
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.
5 0
3 years ago
DO TODAY, PLS HURRY I DONT HAVE TIME :C (10 points, this is science, and its a writeing Question) some animals survive winter by
Lerok [7]

Answer: Beavers

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7 0
3 years ago
An object weighting 9.6 grams is placed in a graduated cylinder displacing the volume from 10.0mL to 13.2 mL
grigory [225]

The volume of object is 3.2 ml

<h3>Explanation:</h3>

Given:

Mass of the object = M = 9.6 g

Initial volume of liquid: V_1 = 10.0\ ml

Final volume of liquid after displacement: V_2 = 13.2\ ml

Total volume of the displaced object inside a graduating cylinder will be given as difference between the final volume and initial volume of the expanding object.

V = V_2-V_1 = 13.2 -1 0.0 = 3.2\ ml

V = 3.2 ml

3 0
3 years ago
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