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monitta
3 years ago
8

What is the lowest common multiple of 4, 10, and 16?

Chemistry
1 answer:
jonny [76]3 years ago
3 0
The least common multiples of 4, 10, and 16 is 80.
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Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.
BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
  • Reading the Periodic Table
  • Writing Compounds
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

<u>Step 3: Use Dimensional Analysis</u>

<u />6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} ) = 511.901 g NO₂

<u>Step 4: Check</u>

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

511.901 g NO₂ ≈ 510 g NO₂

6 0
3 years ago
The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only produ
ohaa [14]
A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67

Then we multiply a number in order to obtain a whole number ratio between the atoms.

1        CH2.67
2       C2H5.34
3       C3H8   <-------- empirical formula
8 0
3 years ago
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