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frutty [35]
3 years ago
13

How many atoms of hydrogen and oxygen are present in 5 gm of HNO3

Chemistry
2 answers:
Ghella [55]3 years ago
7 0

ANSWER: note the amounts of atoms of all the component in HNO3, which are 1 <em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>hydrogen,</u></em><em><u> </u></em><em><u>1</u></em><em><u> </u></em><em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>nitrogen </u></em><em><u>and </u></em><em><u>3</u></em><em><u> </u></em><em><u>atom</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>oxygen</u></em><em><u>.</u></em>

Rufina [12.5K]3 years ago
6 0
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Which ion does not have a noble gas configuration in its ground state?
goldfiish [28.3K]
I think the answer would be Ga3+
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3 years ago
What is the energy of a photon with:<br> a wavelength of 827 nm? What type of radiation is it?
xeze [42]

Explanation:

Given parameters:

 Wavelength of photon  = 827nm  = 827  x 10⁻⁹m

Unknown:

Energy of the photon = ?

Type of radiation = ?

Solution:

The energy of a photon can be derived using the expression below:

      E = \frac{h c}{wavelength}

h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s

c is the speed of light = 3 x 10⁸m/s

Insert the parameters and solve;

      E  = \frac{6.63 x 10^{-34} x 3 x 10^{8}  }{827x 10^{-9} }  

       E = 2.4 x 10⁻¹⁹J

Type of radiation:

Near infrared radiation

 

7 0
3 years ago
What is the mass of 2.0 moles of CO2
igor_vitrenko [27]
1 mol of CO2 is 44.01g/mol
So multiply that by 2 to get 2 mol of CO2, which is 88.02g
6 0
3 years ago
The periodic law states that the physical and chemical properties tend to repeat periodically when elements are arranged accordi
Ivanshal [37]
Atomic number or the number of protons in the element.
6 0
3 years ago
How many molecules of co2 in 97.3 grams of co2
s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

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

Explanation:

<u>Step 1: Define</u>

97.3 g CO₂

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

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.33138 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

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

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

8 0
3 years ago
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