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2 years ago

Hello can you please help me to solve above questions. .

Chemistry

1 answer:

topjm [15]2 years ago

7 0

Answer:

i)a. P & R

b. Q& S

ii) R

iii) Atomic no & Atomic radius

2)a. C4H10

b. CH3Cl

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Avagadros Law..<br>Definition please..

Nikitich [7]

Answer:

"Avogadro's law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The law is a specific case of the ideal gas law. A modern statement is: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

8 0

2 years ago

Read 2 more answers

Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.

OlgaM077 [116]

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

3 0

3 years ago

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Is a burning log an exothermic or endothermic event if the log is the system?

denpristay [2]

It would be endothermic because the log is in the system.

6 0

2 years ago

Please Help! I will give a Brainliest and 18 points! Please do not give me a random, gibberish answer or I will report you and g

3241004551 [841]

Answer:

V2 = 600ml

Explanation:

dilution principle formula

M1V1 = M2V2

C1V1 = C2V2

3 X 20 = 0.1 x V2

60 = 0.1 x V2

V2 = 60/0.1

V2 = 600ml

pls give brainliest

5 0

3 years ago

Read 2 more answers

How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol

SVEN [57.7K]

<h3>Answer:</h3>

0.0157 g Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u /> $3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} )$ = 0.015733 g Au

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.015733 g Au ≈ 0.0157 g Au

3 0

2 years ago

Hello can you please help me to solve above questions. .​

Hello can you please help me to solve above questions. .