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Harman [31]
3 years ago
9

How many moles of NH3 are required to produce 12 moles of NH4Cl

Chemistry
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

16 moles

Explanation:

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Which of the following values are not equal to 1 mole?
Gekata [30.6K]

Answer:

none of them are equal to one mole

3 0
3 years ago
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 353 mL .
vredina [299]
Mass=density·volume. The density is 2.70g/mL and the volume is 353mL. So you would multiply 2.70g/mL by 353mL which will give you 953.1g. Hope that helps :)
7 0
3 years ago
Helppppp plzzz for 30 points
MrRissso [65]

Answer with Explanation:

This is expirament based Q

1) Bring a magnet near ... the Cobalt will come out of te mixture and get attracted to magnet

2)  Disolve it in a solution of ethanol. The Idoine gets dissolve and the other doesnt.

Hope im right!!

5 0
3 years ago
Read 2 more answers
What is the mass in grams of 2.25 mol of the element iron, Fe?
scZoUnD [109]
<h3>Answer:</h3>

126 g Fe

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.25 mol Fe

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

Molar Mass of Fe - 55.85 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2.25 \ mol \ Fe(\frac{55.85 \ g \ Fe}{1 \ mol \ Fe})
  2. Multiply:                            \displaystyle 125.663 \ g \ Fe

<u>Step 4: Check</u>

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

125.663 g Fe ≈ 126 g Fe

4 0
3 years ago
A sample of radioactive material starts with 80 grams. After 3 half-lives have passed, how much radioactive sample remains?
Elanso [62]

Answer:

After 3 half lives 10 g of radio active material left.

Explanation:

Given data:

Total amount of radio active material = 80 g

Amount left after 3 half lives = ?

Solution:

At time zero = 80 g

At first half life = 80 g/2 = 40 g

At 2nd half life = 40 g/2 = 20 g

At 3rd half life = 20 g/2 = 10 g

Thus, after 3 half lives 10 g of radio active material left.

5 0
3 years ago
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