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WINSTONCH [101]
3 years ago
14

Oliver needs to standardize a base solution by titration with an acid solution of known concentration. He finds the volume of th

e acid solution needed to reach the endpoint for a sample of the base solution. Arrange the steps in the order that allows Oliver to determine the concentration of the base solution.
a. Write a balanced equation for the between the analyte and the titrant.
b. Calculate the number of moles of analyte using the stoichiometric coefficients Of the equation.
c. Calculate One number moles of titrant using One volume of titrant required and the concentration of titrant
d. Calculate Ole concentration of the analyte using the number of moles of analyte and the volume of analyte titrated.
Chemistry
1 answer:
Olin [163]3 years ago
7 0
Write a balance equation for the reaction between the analyte and the titrant.
Calculate the # of moles of titrant using the volume of titrant required and the concentration of titrant.
Calculate the # of moles of analyte using the stoichiometric coefficients of the equation.
Calculate the concentration of the analyte using the number or moles of analyte and the volume of analyte titrated.
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Please help me with the questions!
dsp73

Answer:

2500kg/m3

Explanation:

Mass=20g= 20×10^-3kg

Volume=8x10^-6m3

<h2>Now we know that Density = mass\volume</h2>

D= 20×10^-3/8×10^-6

=2500kg/m3

5 0
3 years ago
Substances that cause the oxidation of other substances are
svp [43]
<h3>SUBSTANCE</h3>

==============================

\large  \sf\underline{Question:}

  • Substances that cause the oxidation of other substances are;

==============================

\large  \sf\underline{Answer:}

\qquad \qquad \huge \bold{Option \:  B}

==============================

\large  \sf\underline{Explanation:}

  • Substances that cause the oxidation of other substances are oxidizing agents.

==============================

4 0
2 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
Suppose you take a trip to a distant universe?...
zheka24 [161]
<span>a. the orbital is defined by n,L, mL so (n, L, mL, -1), (n, L,mL, 0) and (n,L,mL, +1) and 3 electrons for any given orbital

b. in (n,L,mL,ms) format the first 12 elements would look like this

(1, 0, 0, +1)
(1, 0, 0, 0)
(1, 0, 0, -1)
(2, 0, 0, +1)
(2, 0, 0, 0)
(2, 0, 0, -1)
(2, 1, 0, +1)
(2, 1, 1, +1)<-----ANSWER
(2, 1, 0, 0)
(2, 1, 1, 0)
(2, 1, 0, -1)
(2, 1, 1, -1)
the idea is we don't pair up electrons until all the mL's have 1 so we wouldn't write
(2, 1, 0, +1)
(2, 1, 0, 0)
(2, 1, 0, -1)
then.
(2, 1, 1, +1)
(2, 1, 1, 0)
(2, 1, 1, -1)
because they would fill
(2, 1, 0, +1)1st
(2, 1, 0, 0)3rd
(2, 1, 0, -1)5th
then.
(2, 1, 1, +1)2nd
(2, 1, 1, 0)4th
(2, 1, 1, -1)6th
to pair (or rather triple up) electrons last
c. ideal gases are when each n level is full...
(1, 0, 0, +1)
(1, 0, 0, 0)
(1, 0, 0, -1)<----- ideal gas 3 electrons so 3 protons and atomic # = 3
(2, 0, 0, +1)
(2, 0, 0, 0)
(2, 0, 0, -1)
(2, 1, 0, +1)
(2, 1, 1, +1)
(2, 1, 0, 0)
(2, 1, 1, 0)
(2, 1, 0, -1)
(2, 1, 1, -1)<----- 2nd ideal gas12 e's so 12 p's and atomic # = 12

continuing on
(3, 0, 0, +1)
(3, 0, 0, 0)
(3, 0, 0, -1)
(3, 1, 0, +1)
(3, 1, 1, +1)
(3, 1, 0, 0)
(3, 1, 1, 0)
(3, 1, 0, -1)
(3, 1, 1, -1)..
(3, 2, 0, +1)
(3, 2, 1, +1)
(3, 2, 2, +1)
(3, 2, 0, 0)
(3, 2, 1, 0)
(3, 2, 2, 0)
(3, 2, 0, -1)
(3, 2, 1, -1)
(3, 2, 2, -1)<--- 3rd nobel gas atomic # = 30
hope it helps
</span>
5 0
3 years ago
Read 2 more answers
With circulation, the heart provides your body with:
Ivanshal [37]
<h3>Answer:  D) all of the above</h3>

Explanation:

The lungs pump oxygen in and carbon dioxide out, which goes through the blood stream to help with the cell's energy needs.

Nutrients pass through the blood stream as well. The nutrients start with the digestive system (mouth, esophagus, stomach, small intestine) before going into the blood stream. The nutrients are building blocks to help make new cells when old ones die off.

When those cells die off, the body sheds them like dead skin, but internal dead cells are passed off as waste. This waste and other byproducts the body doesn't need passes through the blood stream as well.

In short, the blood stream is basically the highway to help get desired materials (eg: oxygen and nutrients) and get rid of waste (eg: carbon dioxide and other unwanted byproducts or dead cell material)

So that's why the answer includes A, B and C.

8 0
3 years ago
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