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vlada-n [284]
3 years ago
6

Find the mass of 1.59 mol of NH3​

Chemistry
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

27.03 grams

Explanation:

moles (n) = mass(m)/Molar mass(M)

moles = 1.59

mass = ?

N = 14, H = 1

M(NH3) = 14 + 3(1) = 17

Substitute the values into the equation:

nM = m

∴ m = 1.59 × 17 = 27.03

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If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?​
alexdok [17]

Answer:

518 mL

Explanation:

We can solve this using Boyle's Law Formula

P1V1 = P2V2

where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume

here , the initial pressure is 1 atm and the initial volume is 725mL

we are given the final pressure 1.4 and we need to find the final volume

so we have p1v1 = p2v2

==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4

(1)(725) = (1.4)v2

==> multiply 1 and 725

725 = (1.4)(v2)

==> divide both sides by 1.4

v2 = 518

N2 would have a volume of 518mL at 1.4atm

7 0
2 years ago
Read 2 more answers
How many molecules are there in 79g of Fe2O3? how many atoms is this?
Leya [2.2K]
There are approximately 160 grams in 1 mol of Fe2O3 molecules. Therefore, there would be 79/160= 0.49375 mols of Fe2O3 molecules in 79 grams. There are 5 atoms in total for each molecule of Fe2O3, therefore 79/160 * 5 = 79/32 = 2.46875 mols of atoms.
4 0
3 years ago
Determine a massa de um átomo de carbono12 em gramas
lapo4ka [179]
I believe it would be 1.660539040 × 10−24 gram.
3 0
3 years ago
Read 2 more answers
What is the mass in grams of 1.00 x 10 24 atoms of Mn?
Alex17521 [72]
<h3>Answer:</h3>

91.2 g Mn

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

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

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                           \displaystyle 91.2321 \ g \ Mn

<u>Step 4: Check</u>

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

91.2321 g Mn ≈ 91.2 g Mn

7 0
3 years ago
Read 2 more answers
How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?
Vsevolod [243]
0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 × 0.35) = 04025 moles
From the formula of molarity moles = molarity × volume in liters
1 mole of KBr is equivalent to 119 g
Therefore, the mass = 0.04025 ×  119  g = 4.79 g
6 0
3 years ago
Read 2 more answers
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