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kompoz [17]
3 years ago
14

How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?

Chemistry
1 answer:
ser-zykov [4K]3 years ago
5 0

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

<em>x</em> = 75/10 = 7.5

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In an experiment, you measure a solution absorbance of 0.2 with a path length of 1cm. If the molar absorptivity coefficient is 5
postnew [5]

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

<h3>What does Beer-Lambert law state?</h3>

The Beer-Lambert law states that for a given material sample, path length and concentration of the sample are directly proportional to the absorbance of the light.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, we can calculate the molarity of the solution using the following expression.

A = ε × b × c

c = A / ε × b

c = 0.2 / (59 cm⁻¹ M⁻¹) × 1 cm = 0.003 M

where,

  • A is the absorbance.
  • ε is the path length.
  • b is the molar absorptivity coefficient.
  • c is the molar concentration.

A solution has an absorbance of 0.2 with a path length of 1 cm. Given the molar absorptivity coefficient is 59 cm⁻¹ M⁻¹, the molarity is 0.003 M.

Learn more about the Beer-Lambert law here: brainly.com/question/12975133

7 0
2 years ago
Question: Draw a valid Lewis structure for the molecule CH3NO in which there are no nonzero formal charges on any of the atoms.
iVinArrow [24]

The mentioned molecule with formula, CH₃NO where no bond is found between N and O can be depicted as formamide is shown in the attachment below.  

The formal charges = Number of valence electrons for the atom (V) - the number of electrons in lone pairs (N) - 1/2 (number of electrons in bond pairs, B)

FC = V - N - B/2

Thus, there is a need to calculate valence electrons, electrons in lone pairs, and the number of electrons in bond pairs for each atom in the mentioned molecule.  

V or valence electrons on C = 4e, on H = 1e, on N = 5e, and on O = 6e.  

N or electrons in lone pairs on C = 0e, on H = 0e, on N = 2e, and on O = 4e.  

B or number of electrons in bond pairs for C = 8e, for H = 2e, for N = 6e, and for O = 4e.  

Thus, the formal charges for each will be,  

C = 4 - 0 - (8/2) = 0

H = 1 - 0 - (2/2) = 0

N = 5 - 2 - (6/2) = 0

O = 6 - 4 - (4/2) = 0

Lewis dot structure for the given molecule is given in the attachment below:


3 0
3 years ago
Read 2 more answers
36g of an alloy of copper and zinc contains 45% copper. how much pure copper do you need to add in order to get a 60% copper all
iogann1982 [59]
First, you need to count copper mass in alloy.
Second, you have to make an equation an find x ( the copper mass must be added). The answer is: 13,5g pure copper

7 0
3 years ago
Use the solubility generalizations on the information page to predict if one or more precipitates will form when aqueous solutio
neonofarm [45]

Answer:

3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)

Explanation:

In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>

The reaction is:

<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>
6 0
3 years ago
at do oxygen (O) and bromine (Br) have in common? They both tend to gain electrons to gain stability. They both have seven valen
avanturin [10]
Both of them have high electronegativity. Hence they both tend to gain electrons to gain stability.
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