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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so

Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}$

$\text{Molar mass of unknown substance}=39.7g/mol$

Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0

2 years ago

Metals react with ______ to form compounds that are alkaline.

kondor19780726 [428]

Metals react with ______ to form compounds that are alkaline.

A. metalloids

B. oxygen (O)

C. non-metals

D. hydrogen (H)

The answer is D, Hydrogen (H).

7 0

2 years ago

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Fossils found in layers of sediment give us clues to Earth's past. Based on the fossil record of this area, we can assume that t

Shalnov [3]

Answer:

its b

Explanation:

i took the test

8 0

2 years ago

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The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol

Dafna11 [192]

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂

The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:

15.1 mol × 44.01 g/mol = 665 g

4 0

2 years ago

Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how

Aleonysh [2.5K]

Given equation : $5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O$

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

$Moles = \frac{grams}{molar mass}$

Molar mass of KNO2 = 85.10 g/mol

$Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}$

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

$0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}$

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

$Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}$

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

$2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}$

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0

2 years ago