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snow_tiger [21]
4 years ago
7

A sample of an unknown compound contains 0.350 moles of sulfur and 1.05 moles of oxygen. The molecular mass of the compound is 2

40.18 g/mol. What is the molecular formula of this compound?
Chemistry
1 answer:
jeyben [28]4 years ago
4 0

Answer:

true

Explanation:

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CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?
Kipish [7]
<h3>Answer:</h3>

64 g O₂

<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

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0
3 years ago
Read 2 more answers
How does the density of a gas depend on the molar mass of the gas?
AlladinOne [14]

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

\rho = \frac{m}{V}                          ......equation (1)

According to the ideal gas law for ideal gas:

PV = nRT                                       ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: n = \frac{m}{M}

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

PV = \frac{m}{M}RT

⇒ PM= \frac{m}{V} RT

⇒ \frac{PM}{RT}= \frac{m}{V}             ......equation (3)

Now from equation (1) and (3), we get

\frac{PM}{RT}= \frac{m}{V} = \rho  

⇒ Density of an ideal gas: \rho = \frac{PM}{RT}  

⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>

<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>

6 0
4 years ago
How many valence electrons are in the element gallium? explain how you determined the number.
Natasha2012 [34]

Answer:

three valence electrons

Explanation:

Gallium has three electrons in the outer energy level and therefore has three valence electrons. The identification of valence electrons is vital because the chemical behavior of an element is determined primarily by the arrangement of the electrons in the valence shell.

4 0
4 years ago
Science plz help due soon 6th grade Major grade
pentagon [3]

Answer:

the answer is d.

Explanation:

because the core is deepest then is the mantle and the lithpsphere is last .

I put a picture incase you need a visual example.

5 0
3 years ago
if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti
ehidna [41]
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
7 0
4 years ago
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