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aliya0001 [1]
3 years ago
5

How many moles are found in a 350.mL solution of 13.5M HNO3

Chemistry
1 answer:
tiny-mole [99]3 years ago
5 0

Answer: Moles are found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

Explanation:

Given: Volume = 350 mL (1 mL = 0.001 L) = 0.350 L

Molarity = 13.5 M

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

Therefore, moles of HNO_{3} present in the given solution are calculated as follows.

Molarity = \frac{moles}{Volume (in L)}\\13.5 M = \frac{moles}{0.350 L}\\moles = 4.725 mol

Thus, we can conclude that moles found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

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Where on the table would a group's<br> ionization energy be the greatest?
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Answer:

upper left

Explanation:

it is a trend on the periodic table. Ionization energy increases from left to right(->) I t also increases down to up

5 0
3 years ago
A graduated cylinder was filled to 20.3 mL with water. A solid object weighing 73.05 g was immersed in the water, raising the me
victus00 [196]

The density of the solid object will be 2.63 g/mL

<h3>What is density?</h3>

Density of objects = mass/volume.

Recall that an object will always displace its own volume when placed in a liquid.

Volume of the solid object = Cylinder reading after immersing the object in the water - cylinder reading before immersing the object in the water.

             = 48.1 - 20.4

                  = 27.8 mL

Mass of the solid object = 73.05 g

Density of the object = 73.05/27.8

                                         = 2.63 g/mL

More on density can be found here: brainly.com/question/15164682

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3 0
10 months ago
Which atom would it be most difficult to remove an electron from
Norma-Jean [14]
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements).  This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove.  Examples Helium, Neon, Argon, Xenon, Krypton and Radon 
6 0
3 years ago
How many moles are in 1.5L of 0.40M Na2SO4?​
anzhelika [568]

Answer:

0.60 moles of Na₂SO₄

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

A solution of Na₂SO₄ 0.40M contains 0.40 moles of solute (Na₂SO₄) per liter of solution.

As you have 1.5L of solution, moles of Na₂SO₄:

1.5L × (0.40mol / L) = <em>0.60 moles of Na₂SO₄</em>

7 0
3 years ago
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

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

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
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