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Neko [114]
2 years ago
5

Which element has a higher first ionization energy than chlorine (Cl)?

Chemistry
2 answers:
4vir4ik [10]2 years ago
6 0

Answer:

Argon

Explanation:

It has more electron than chlorine

erastova [34]2 years ago
5 0
Argon- because if you look at the periodic table- as you move left from right the IE will increase
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One liter of air contains about 0.21 L of oxygen. When filled, the human lungs hold about 6.0 L of air. How much oxygen is in th
masya89 [10]
Since we are told that 1L of air contains 0.21L of oxygen, you can use the conversion (0.21L O₂)/(1L air).  That means that you can just multiply 6.0L by 0.21L to get 1.26L of O₂.
that means that the lungs can hold about 1.26L of oxygen.
I hope this helps.  Let me know if anything is unclear.
4 0
3 years ago
What is the atomic structure for antimony
djverab [1.8K]

Answer:

Hmm

Explanation:

Elemental antimony adults a layer structure (space group R3m No.166) in which layers consist of fused, ruffled, six-membered rings. The nearest and the next nearest neighbors form and irregular octahedral complex, with the 3 atoms in each layer slightly closer than 3 atoms in the next

Electric configuration : 4d^10 5s^2 5p^3

3 0
2 years ago
N2H4 + N2O4 --> N2 + H2O
ahrayia [7]

Answer:

  1. The limiting reagent is N2O4
  2. 14,09g

Explanation:

  • First, we adjust the reaction.

2N_{2} H_{4} + N_{2} O_{4} ⇄6N_{2} +  4H_{2}O

  • Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using N_{2} H_{4} to form H_{2}O

               molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4}  }{32,04\frac{g}{mol}  N_{2} H_{4} }

                                           molH_{2} O = 1, 125 mol

Using N_{2} O_{4} to form H_{2} O

              molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4}  }{92\frac{g}{mol}  N_{2} O_{4} }

                                           molH_{2} O = 0,783 mol

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴                          MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}

                                      MassOfH_{2}O = 14,09g

5 0
3 years ago
How many grams of Nacl would you need to prepare 200 ml of a 5 M SOLUTION. ?
vazorg [7]

Answer:

58.44 g of NaCl are needed.

Explanation:

Given data:

Mass of NaCl needed = ?

Volume of solution = 200 mL (200/1000 =0.2 L)

Molarity of solution = 5 M

Solution:

We will solve this problem through molarity formula.

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will put the values.

5 M = moles of solute / 0.2 L

Moles of solute = 5 mol/L × 0.2 L

Moles of solute = 1 mol

Mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 1 mol × 58.44 g/mol

Mass = 58.44 g

Thus, 58.44 g of NaCl needed.

8 0
3 years ago
Can someone help me solve this please 2.150 x 10 ^-12 dm / 5.02 x 10 ^-5dm^3
zavuch27 [327]

4.3 x 10⁻⁸dm⁻²

Explanation:

Problem:

                   \frac{2.150 x 10^{-12} dm }{5.02  x 10 ^{-5} dm^{3} }

 To solve this problem, we have apply the right rule of indices operation.

If we have an indices in the form;

 \frac{a^{n} }{b^{k} } = (\frac{a }{b } )^{n-k}

  Solving this problem:

   = \frac{2.15}{5.02}   x  10^{-12 -(-5})

   = 0.43 x 10⁻⁷

   = 4.3  x 10⁻¹ x 10⁻⁷

   = 4.3 x 10⁻⁸

for the units:

 \frac{dm}{dm^{3} }   = \frac{dm}{ dm x dm x dm} = \frac{1}{dm^{2} }  = dm^{-2}

  The solution is:  4.3 x 10⁻⁸dm⁻²

   

learn more:

long division brainly.com/question/1747117

#learnwithBrainly

4 0
3 years ago
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