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3 years ago

The final volume of the solution is 284 mL. What is the concentration of CuSO4 in the final solution, in mol/L?

Chemistry

1 answer:

DochEvi [55]3 years ago

7 0

Answer:

0.0252mol/L

Explanation:

The following data were obtained obtained from the question:

Volume of solution = 284mL = 284/1000 = 0.284L

Mole of CuSO4 = 7.157 × 10^-3 mol

Molarity =?

Molarity = mole/Volume

Molarity = 7.157x10^-3 /0.284

Molarity = 0.0252mol/L

The concentration of the solution is 0.0252mol/L

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NEED ASAP

notka56 [123]

Answer: B

Refer:

Because water molecules are polar, they interact with the sodium and chloride ions. In general, polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes. This concept is often expressed as “Like dissolves like.

5 0

2 years ago

In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i

GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: $V_{1}$ = 21.35 mL, $M_{1}$ = 0.150 M

$V_{2}$ = 25.0 mL, $M_{2}$ = ?

Formula used to calculate molarity of $H_{2}SO_{4}$ is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M$

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = $V_{1} + V_{2}$

= 21.35 mL + 25.0 mL

= 46.36 mL (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol$

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0

2 years ago

15. When traveling through a smoke-filled area, a person must be able

maxonik [38]

When traveling through a smoke-filled area, a person must be able to have distance of at least 4 feet in order to safely reach a fire exit. That is option C.

<h3>What is a smoke-filled area?</h3>

Smoke-filled area is an enclosed area that is filled with smoke gas that is usually made up of carbon monoxide.

The smoke-filled area is usually caused by

Fire out break

Burning of plastics or chemical products

Release of fuming gases.

When there is smoke-filled area die to fire out break, there are some guidelines to follow to safely reach the fire exit early enough.

The importance of these guidelines include the following:

enable rescuers to navigate a scene more quickly,

identify risks and hazards,

locate safety points,

determine the safest way in and out of a building, and

map out evacuation routes.

The affected individual should get down and crawl while taking short breath through the nose because cleaner air is nearest to the floor.

Learn more about distance here:

brainly.com/question/7243416

#SPJ1

5 0

1 year ago

Calculate the number of atoms in 0.25 moles of manganese?

White raven [17]

<h3>Answer:</h3>

1.5 × 10²³ atoms Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.25 mol Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.25 \ mol \ Mn(\frac{6.022 \cdot 10^{23} \ atoms \ Mn}{1 \ mol \ Mn})$
Multiply: $\displaystyle 1.5055 \cdot 10^{23} \ atoms \ Mn$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.5055 × 10²³ atoms Mn ≈ 1.5 × 10²³ atoms Mn

3 0

2 years ago

In lab an experiment was performed to determine the empirical formula of magnesium

Nat2105 [25]

Answer:

The correct option is

C) Trial 1 will have the same calculated empirical formula as trial 2.

Explanation:

The empirical formula is the formula of a chemical compound that states the simplest whole number ratio of each of the atoms included in the compound. It is obtained by dividing the mass of an element present in the compound by the element's molar mass to find the mole ratio of the elements. The obtained mole value for each element is then divided by the smallest number of moles obtained in the division.

By definition the composition and ratio of elements combined in a chemical compound is fixed, therefore trial 1 will have the same calculated empirical formula as trial 2.

5 0

2 years ago