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puteri [66]
3 years ago
15

PLEASE HELP ME IM SO LOST

Chemistry
1 answer:
fenix001 [56]3 years ago
7 0
I am stuck on this too
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Which of the following elements is likely to show additional stability in its electronic configurations because of filled and ha
kodGreya [7K]
Ah is gold and Rh is rhodium and Os is osmium and Ir iridium
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3 years ago
Why does green light slow down more than orange light does when passing through an object?
otez555 [7]
The correct answer among the choices is the last option. <span>Green light has a higher frequency than orange light. Frequency is inversely proportional to wavelength. G</span>reen has wavelengths ranging form 495 to 570 nm and orange has wavelengths ranging from 590 to 620 nm. Speed is said to be directly proportional with wavelength. Higher wavelength means more faster light.
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4 years ago
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How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?
Maru [420]

When sodium amide i.e.NaNH_{2} reacts with water i.e. H_{2}O results in the formation of sodium hydroxide i.e. NaOH and ammonia NH_{3}.

The chemical reaction is given by:

NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH

Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

The chemical reaction is given by:

NH_{3}+H_{2}O\rightarrow NH_4OH

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


8 0
3 years ago
50 points plz help
EleoNora [17]

Answer: b

Explanation:

7 0
3 years ago
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