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Vedmedyk [2.9K]
2 years ago
8

A measure of the strength of acid rain can be given using _____.

Chemistry
1 answer:
Gre4nikov [31]2 years ago
8 0
The pH scale is used to measure the strength of acids and bases.
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Why is the anatomy of the human body not a general reason to study chemistry. Need help for a good reply.
ivolga24 [154]
Anatomy's a biology topic!
8 0
2 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
Pls that is the picture ​
Fiesta28 [93]
Do what it says easy
3 0
2 years ago
A cylindrical glass tube of length 27.4cm and radius 4 cm is filled with gas. The empty tube has a mass of 254.3 g. The tube fil
Temka [501]

Density of the gas is 3.05 × 10⁻³ g / cm³.

<u>Explanation:</u>

Volume of the cylinder = π r² h

where r is the radius and h is the height of the height or the length of the glass tube.

Here r = 4 cm and h = 27.4 cm

Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³

We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.

Mass of the substance = 258.5 - 254.3 = 4.2 g

We have to find the density using the formula as,

$ Density = \frac{mass}{volume}

Plugin the values as,

$ Density = \frac{4.2}{1376.6}

              = 3.05 × 10⁻³ g / cm³

So the Density of the gas is 3.05 × 10⁻³ g / cm³.

3 0
2 years ago
A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is
NemiM [27]
The silver chloride electrode usually functions as a redox electrode where the equilibrium is achieved between silver and its salt (silver chloride).
The half reaction for this electrode is as follow:
<span>AgCl(s)+e−→Ag(s)+Cl−(aq) where:
</span>(s) refers to solid state
(aq) refers to the aqueous state and 
e- is the electron
5 0
3 years ago
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