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BARSIC [14]
3 years ago
9

Atoms and bonds are often represented in chemical structures by balls and sticks, but like organelles, are more complex than the

y are often represented. Later, you’ll learn about bonds. For now, you’ll explore what an atom is.
Chemistry
1 answer:
likoan [24]3 years ago
5 0

Answer:

where is the question?

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Mass (g) 4.25 g 7.48 g Initial Volume of Graduated Cylinder (mL) 7 mL 6 mL Final Volume of Graduated Cylinder (mL) 7.5 mL 7 Obje
nata0808 [166]

Answer:

A table was attached to the question

Explanation:

The step by step calculation is as shown

7 0
4 years ago
Which option is an example of a chemical property?
bulgar [2K]
Reactivity
Hope it helps!
7 0
4 years ago
Read 2 more answers
What is the pH of an aqueous solution of 1 M CH3COOH (pKa=4)
Mrac [35]

Answer:

pH = 2.

Explanation:

A weak acid is in equilibrium with its ions in a solution, so it must have an equilibrium constant (Ka). And, pKa = -logKa

Ka = 10^{-pKa}

Ka = 10⁻⁴

So, for CH₃COOH the equilibrium must be:

CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)

1 M                           0                0                Initial

-x                              +x              +x               Reacted

1-x                             x                x                 Equilibrium

And the equilibrium constant:

Ka = \frac{[H+]x[CH3COO-]}{[CH3COOH]}

10^{-4} = \frac{x^2}{1-x}

Supposing x << 1:

10⁻⁴ = x²

x = √10⁻⁴

x = 10⁻² M, so the supposing is correct.

So,

pH = -log[H⁺]

pH = -log10⁻²

pH = 2

8 0
4 years ago
A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol
Vlada [557]

<u>Answer:</u> The mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

<u>Explanation:</u>

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, \frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for 9.4625\times 10^7L of water, the amount of chlorine applied will be \frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg

Hence, the mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

6 0
3 years ago
Question (a).....................
Rudiy27
I dont get the question ???
6 0
3 years ago
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