answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Arlecino [84]
2 years ago
15

A solution contains 25 grams of KNO3 dissolved in 200. grams of H2O. Which numerical setup can be used to calculate the percent

by mass of KNO3 in this solution?
Chemistry
2 answers:
inna [77]2 years ago
4 0

Answer:

25g/225g X 100

Explanation:

% of Mass = [ mass of part ( solute ) / mass of whole ( solution )] x 100

Furkat [3]2 years ago
3 0

Answer:

% = 11.11%

Explanation:

To get the %m/m of any solution we should use the following expression:

%m/m = m solute / m solution * 100

we have the mass of solute, but not the mass of solution, however this can be calculated. solution is made using solute and solvent so:

m solution = 25 + 200 = 225 g

Now that we have the mass of solution, we can calculate the %:

%m/m = 25 / 225 * 100

%m/m = 11.11%

This is the %m/m of this solution

You might be interested in
Is utensils a substance homogeneous mixture or heterogeneous mixture
Phoenix [80]

A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

7 0
2 years ago
At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature
Alja [10]

Answer:

Final pressure = 2.3225 atm

Amontons’s law states that

At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature

Explanation:

Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure

Amontons’s law: P/T = Constant at constant V and n

That is P1/T1 = P2/T2

Where temperature is given in Kelvin

Hence T1 of 10°C = 273.15 + 10 = 283.15K

Also temperature T2 of 40°C = 313.15 K

Hence

P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm

3 0
2 years ago
A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta
GalinKa [24]
Answer:

<span>23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. </span>

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula 

number of moles = mass in grams / molar mass


CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

3 0
2 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
2 years ago
Read 2 more answers
Other questions:
  • How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
    7·1 answer
  • A reagent bottle on the shelf labeled 0.5m nacl was used in place of the 0.5 m cacl2. assuming c2o4^2- to be in excess, what wou
    11·1 answer
  • Of the elements: b, c, f, li, and na. the element with the highest ionization energy is
    5·1 answer
  • Which of the following does not involve colligative properties?
    9·2 answers
  • Two liquids are known to be miscible. They are placed in a beaker and shaken. Which of the following describes what will happen?
    5·2 answers
  • A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft
    7·2 answers
  • 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu
    9·1 answer
  • An athlete has 15% body fat by mass. What is the weight of fat, in pounds, of a 74 kg-athlete?
    13·1 answer
  • The engineers at a hydroelectric dam designed a computer program that will allow them to determine the amount of energy they can
    10·1 answer
  • How many hydrogen molecules are there in 1 ton of hydrogen?​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!