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Katyanochek1 [597]
2 years ago
11

L 6.3.2 Test (CST):

Chemistry
1 answer:
dalvyx [7]2 years ago
5 0

B. Water Temperature {A PE X}

Explanation:

You might be interested in
The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.
Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0
3 years ago
No I don’t have a question
Nikitich [7]

Answer:

then why'd you ask a question

Explanation:

3 0
2 years ago
In a exothermic reaction, is heat transfered to or from the surrounding
Monica [59]
In an exothermic reaction, heat is transferred to the surrounding.

Hope this helps, have a great day ahead!
7 0
3 years ago
Read 2 more answers
A solution is made containing 4.6 g of sodium chloride per 250g of water.
lozanna [386]

Answer:

\large \boxed{1.81 \, \%}

Explanation:

\text{Mass percent} = \dfrac{\text{mass of solute} }{\text{mass of solution}} \times 100 \, \%

Data:

Mass of NaCl =      4.6 g

Mass of water = 250    g

Calculations:

Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.

\text{\% m/m} = \dfrac{\text{4.6 g} }{\text{254.6 g}} \times 100 \, \% = \mathbf{1.81 \, \%}\\\\\text{The percent by mass of NaCl is $\large \boxed{\mathbf{1.81 \, \%}}$}

3 0
3 years ago
Compare the volume of 14.1 g of helium to 14.1 g of argon gas (under identical conditions).
s2008m [1.1K]
The Volumes can be calculated from Masses by using following Formula,

                                        Density  =  Mass / Volume
Solving for Volume,
                                        Volume  =  Mass / Density

Mass of Both Gases  =  14.1 g

Density of Argon at S.T.P  =  1.784 g/L

Density of Helium at S.T.P  =  0.179 g/L

For Argon:
                                        Volume  =  14.1 g / 1.784 g/L

                                        Volume  =  7.90 L

For Helium:
                                        Volume  =  14.1 g / 0.179 g/L

                                        Volume  =  78.77 L
4 0
3 years ago
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