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ollegr [7]
3 years ago
5

Which of these periods contain elements with electrons in s, p, d, and forbitals?

Chemistry
2 answers:
yarga [219]3 years ago
7 0

Answer:

The periods that would contain elements with electrons in s, p, d, and f orbitals are periods 6-7

Explanation:

vodka [1.7K]3 years ago
5 0

Answer:

I learnt k,l,m

Explanation:

so 1-3periods have klm and after that klmn

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The formula is SrCl2. hope this helps
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Aqueous hydrobromic acid hbr will react with solid sodium hydroxide naoh to produce aqueous sodium bromide nabr and liquid water
Natalka [10]
 Balanced equation is

HBr + NaOH ----> NaBr + H2O

Using molar masses

80.912 g HBr reacts with  39.997 g of Naoh to give 18.007 g water

so 1 gram of NaOH reacts with 2.023 g of HBR   
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction

Mass  of water produced  =    (5.7 * 18.007 / 39.997  =  2.6 g to 2 sig figs
8 0
3 years ago
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You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl
bulgar [2K]

Answer: Mass of CO_2  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)

Mass or reactants =  Mass of CaCO_3+ mass of HCl = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of CO_2 + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of CO_2  produced in this reaction was 6.56 grams

7 0
3 years ago
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Molar mass of butane:-

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Moles of Butane:-

  • 100/58=1.7mol

#16

  • 2mols of Butane produce 8mol CO_2
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Moles of CO_2

  • 4(1.7)=6.8mol

Mass of CO_2:-

  • 44(6.8)=299.2g

#17

  • 2mols of butane need 13mol O_2
  • 1mol of butane needs 6.5mol O_2

Moles of O_2

  • 6.5(1.7)=11.05mol

Mass of O_2

  • 11.05(32)=353.6g
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