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serious [3.7K]
3 years ago
6

Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi

xed as thoroughly as possible what would be the density of the resulting mixture?
Chemistry
1 answer:
KiRa [710]3 years ago
6 0

The density of the mixture will be 1.57 g/cm³.


Step 1. Calculate the <em>mass of the butter</em>.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2. Calculate the <em>mass of the sand</em>.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3. Calculate the <em>density of the mixture</em>.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

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lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

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Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

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NaOH

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a

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So your solution has

[

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=

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Now, the

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[

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−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

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−

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∘

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14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

3 0
2 years ago
1. A student attempts to make a saline solution by adding salt to water. How much water did
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111.1 mL of water

Explanation:

Weight per volume concentration (w/v %) is defined as

weight per volume concentration = (mass of solute (g) / volume of solution (mL)) × 100

volume of solution = (mass of solute × 100) / weight per volume concentration

volume of solution = (1 × 100) / 0.9 = 111.1 mL

volume of water = volume of solution = 111.1 mL

Learn more about:

weight per volume concentration

brainly.com/question/12721794

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8 0
3 years ago
What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed
Ahat [919]

Answer:

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Solution:

Data Given;

                   Mass  =  84.7 g

                   Volume  =  49.6 cm³

                   Density  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Putting values,

                   Density  =  84.7 g ÷ 49.6 cm³ 

                   Density  =  1.70 g.cm⁻³

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How many liters of a 4.0 M CaCl2 solution would contain 2 moles of CaCl2?
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Answer:

The answer to your question is 0.5 liters

Explanation:

Data

[CaCl₂] = 4.0 M

number of moles = 2

volume = ?

Process

To solve this problem use the formula of Molarity and solve it for volume, substitute the values and simplify.

-Formula

Molarity = moles / volume

-Solve for volume

Volume = moles / molarity

-Substitution

Volume = 2/4

-Simplification

Volume = 0.5 liters.

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3 years ago
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solar is better than wind

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7 0
2 years ago
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