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3 years ago

How many atoms are in 3.05 grams of copper?

Chemistry

1 answer:

vfiekz [6]3 years ago

5 0

6.02 x10^23 atom

3.5g x 1mol/63.55g Cu x 6.02 x 10^23/ 1mol=

3.32 x 10^22 atoms

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What is the charge on an electron?

Llana [10]

Answer:

The answer is negative .

3 0

2 years ago

How many grams of HF are needed to react with 3.0 moles of Sn?

Flauer [41]

Answer:

120g

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction.

Sn + 2HF —> SnF2 + H2

Step 2:

Determination of the number of mole HF needed to react with 3 moles of Sn.

From the balanced equation above,

1 mole of Sn and reacted with 2 moles of HF.

Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.

Step 3:

Conversion of 6 moles of HF to grams.

Number of mole HF = 6 moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn.

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2 years ago

What are the products obtained in the electrolysis of molten nai?

yanalaym [24]

Answer is: sodium (Na) and iodine (I₂).

First ionic bonds in this salt are separeted because of heat:
NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.

4 0

2 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]

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

Explanation:

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

2 years ago

Answer both 8 and 9 will give u brain list with explanation as well so don’t just answer

goldfiish [28.3K]

Answer:

8. the answer is B.

9. the answer is A.

Explanation:

Hello!

8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

$3molO_2*\frac{2molH_2O}{1molO_2}=6molH_2O$

Thus, the answer is this case is A.

Best regards!

6 0

2 years ago