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Bess [88]
4 years ago
10

How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas

s solution? Show all of the work needed to solve this problem.
Chemistry
1 answer:
IrinaVladis [17]4 years ago
4 0

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

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A nucleotide consists of a phosphate group, a pentose sugar, and a __________________, all linked together by covalent bonds. po
Genrish500 [490]
The correct answer would be the fourth option. A nucleotide consists of a phosphate group, a pentose sugar, and a nitrogen containing base that are all linked together by covalent bonds. Nucleotides are the monomer units of nucleic acids and is the basic unit of the DNA.
5 0
3 years ago
What is the molar concentration of
Leto [7]

Answer:

1.42 M

Explanation:

First calculate the amount of moles.

that's done by dividing the mass with the molecular mass so 660g / 310.18 g/mol = 2.13 mol

Then you can calculate the molarity by dividing the moles with the volume so 2.13 mol / 1.5 l = 1.42 M

(without rounding: 1.418531175 M)

3 0
2 years ago
When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o
Angelina_Jolie [31]

19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.

Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2

Volume of O_2 (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of O_2 (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2C_4H_10 + 13O_2 ---> 8CO_2 + 10H_2O

From reaction;

13 moles O_2 require 2 moles C_4H_10

So, 1.663 moles O_2 will require = 2 x 1.663 ÷13 = 0.256 moles of C_4H_10

Thus C_4H_10 is a limiting reagent. So it will drive the yield of CO_2.

Moles of CO_2 produced = (8/2) 0.2 = 0.8 moles

Pressure of CO_2 (P) = 102 - 2.24 = 99.76 kPa = 99.76  ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for CO_2,

PV = nRT

0.985 V = 0.8 0.0821 x 295

V = 19.7 liter

The volume of CO_2 produced = 19.7 liter.

Learn more about the vapour pressure here:

brainly.com/question/25699778

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5 0
2 years ago
It takes to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond
Zolol [24]

The given question is incomplete. The complete question is :

It takes 151 kJ/mol to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon. Be sure your answer has the correct number of significant digits.

Answer:  793 nm

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

E = energy of the light  = 151 kJ= 151000 J   (1kJ=1000J)

N= moles = 1 = 6.023\times 10^{23}

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light  = 3\times 10^8m/s

\lambda = wavelength of light  = ?

Putting in the values:

151000J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda}

{\lambda}=7.93\times 10^{-7}m=793nm    1m=10^{-9}nm

Thus  the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon is 793 nm

3 0
3 years ago
Which orbital do potassium's outermost electrons occupy? 4s 4p 4d 3s
Elanso [62]
Potassium outermost electron occupy "4s" orbital
4 0
4 years ago
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