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Harlamova29_29 [7]
2 years ago
13

It takes 38mL of 0.75M NaOH solution to completely neutralize 155 mL of a chloric acid solution (HclO3). what is the concentrati

on of the HclO3 solution? Please explain well
Chemistry
1 answer:
Scrat [10]2 years ago
8 0
ITS ABOUT DRIVE
ITS ABOUT POWER
WE STAY HUNGRY
WE DEVOUR
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Which atom would it be most difficult to remove an electron from
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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 
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3 years ago
Sodium carbonate (Na2CO3) reacts with acetic acid (CH3COOH) to form sodium acetate (NaCH3COO), carbon dioxide (CO2), and water (
Evgen [1.6K]

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=3,723.3 J=3.72  but if it has the option of -3.72 kJ then use that

Explanation:

Use the formula q=m×Cp×delta T

m=1.500 kg=1,500 g

Co=2.52 J/g·k

delta T=0.985k

q=(1,500g)(2.52 J/g·k)(0.985k)

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3 years ago
If a quadrilateral has three angels measuring 60, 45 and 100 find the measure of the fourth angel
Nina [5.8K]
The sum of angels of a quadrilateral is equal to 360°.

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7 0
3 years ago
A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the
Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047  mol/dm³

The concentration of the sulfuric acid is 0.047  mol/dm³

Download docx
7 0
2 years ago
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