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Delicious77 [7]
2 years ago
6

Rank the following in terms of increasing ionization energy: Sulfur, oxygen, hydrogen, and fluorine.

Chemistry
1 answer:
Naily [24]2 years ago
3 0

Answer:

Ranked from lowest to highest:

Sulfur - 999.6 kJ/mol

Hydrogen - 13.5984 kJ/mol

Oxygen - 13.6181 kJ/mol

Fluorine - 17.4228 kJ/mol

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COCI2 has an effusion rate of 0.00172 m/sec. Which of the gases below would have an effusion rate of 0.00323 m/sec?
geniusboy [140]
CO por qué si y punto, chao
8 0
2 years ago
When the potential difference across a lamp is 3.3V the current is 0.15 calculate the resistance of the lamp and give the unit
marissa [1.9K]

Answer:

22Ω

Explanation:

Given parameters:

Potential difference  = 3.3V

Current  = 0.15A

Unknown:

Resistance  = ?

Solution:

According to ohm's law, potential difference, current and resistance are related by the expression below;

             V = I R

where V is the voltage

           I is the current

          R is the resistance

          3.3 = 0.15 x R

             R = \frac{3.3}{0.15}   = 22Ω

6 0
2 years ago
Calculate the amount in grams of Na2CO3 needed in a reaction with HCl to produce 120g NaCl?
klemol [59]
The balanced chemical reaction is written as :

Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:

120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
8 0
3 years ago
What volume of water has the same mass as 9.0m3 of ethyl alcohol?
Triss [41]
This question requires the knowledge of density. 

The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³

Density = Mass / Volume 

By applying ethyl alcohol,
    789 kg m⁻³ = Mass / 0.9 m³
         Mass     = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.

Then by applying water,
     1000 kg m⁻³ = 710.1 kg / Volume
      Volume        = 0.7101 m³
                          = 0.7 m³
hence the equal water volume is 0.7 m³
8 0
3 years ago
How many moles of Chromium is in 4.41 ×10^24 atoms
zalisa [80]

7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

<h3>How to find the number of moles ?</h3>

Number of moles = \frac{\text{Given number of atoms}}{\text{Avogadro's Number}}

     

<h3>What is Avogadro's Number ?</h3>

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.

Avogadro's Number = 6.022 × 10²³

Now put the values in above formula we get

Number of moles = \frac{\text{Given number of atoms}}{\text{Avogadro's Number}}

                             = \frac{4.41 \times 10^{24}}{6.022 \times 10^{23}}

                             = 7.32 moles

Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

Learn more about the Avogadro's Number here: brainly.com/question/1581342

#SPJ1

3 0
1 year ago
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