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Greeley [361]
3 years ago
10

What is a dependent variable

Chemistry
2 answers:
Juliette [100K]3 years ago
7 0
The dependent variable is the variable that changes based on the independent variable (the manipulated variable).
SCORPION-xisa [38]3 years ago
6 0
Values of dependent variables depend on the values of independent variables. The dependent variables represent the output or outcome whose variation is being studied.
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A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill
nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<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)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

5 0
3 years ago
Lab Scenario D:
Cloud [144]

Answer:

See explanation

Explanation:

A reaction in which heat and light are produced is a combustion reaction. Combustion is said to have occurred when a substance is burnt in oxygen.

The balanced equation of the reaction is;

4Li(s) + O2(g) ------->2Li2O(s)

This reaction is exothermic because heat was produced. The reaction has a low activation energy as the metal easily burst into flames in oxygen. A catalyst is not needed in this reaction because it has a low activation energy.

According to the law of conservation of mass. Atoms are neither created nor destroyed in a chemical reaction. What this means is that in a chemical reaction, the number of atoms of each element on the left hand side must be the same as the same as the number of atoms of the same element on the right hand side.

3 0
3 years ago
If a car rolls down a ramp, what force is beating what force ?
Katena32 [7]
Well.. if ur going down a ramp, friction beat gravity, right....??? I guess, I I have an idea hahha
8 0
3 years ago
Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry
Ipatiy [6.2K]

<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm

<u>Explanation:</u>

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT

where,

\pi = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of sucrose = 3.40 g

M_{solute} = molar mass of sucrose = 342.3 g/mol

V_{solution} = Volume of solution = 1 L

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 20^oC=[20+273]K=293K

Putting values in above equation, we get:

\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm

Hence, the pressure that must be applied to the apparatus is 0.239 atm

3 0
3 years ago
A solution is made containing 14.6g of CH3OH in 185g H2O.1. Calculate the mole fraction of CH3OH.2. Calculate the mass percent o
Andre45 [30]

Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O

Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

Next, mass percent is:

\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

Regards.

7 0
3 years ago
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