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My name is Ann [436]
2 years ago
11

What is the atomic number of carbon​

Chemistry
1 answer:
rosijanka [135]2 years ago
8 0
Carbon (from Latin: carbo "coal") is a chemical element with the symbol C and atomic number 6. It is nonmetallic and tetravalent—making four electrons available to form covalent chemical bonds. It belongs to group 14 of the periodic table.
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When light strikes chlorophyll molecules, they lose electrons, which are ultimately replaced by other electrons obtained by ____
fomenos
The answer is by "splitting water".

hope this helps
5 0
3 years ago
Which of the following describes a chemical property of oxygen
monitta

Which of the following describes a chemical property of oxygen

8 0
2 years ago
Most bones are connected at Joints by​
Anuta_ua [19.1K]

Answer:

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7 0
3 years ago
Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One of them is a blue solid that comes apa
Marizza181 [45]

Work out the number of moles in

100.00 grams of the oxide.

For nitrogen: The atomic mass of N is 14.0067, and we have 36.84 g N:

36.84 g N14.0067 g N/mol N=2.630 mol N

For oxygen: The atomic mass of O is

15.9994, and we have

100.00−36.84=63.16 g O:

63.16 g N 15.9994 g N/mol N=3.948 mol N

Now the ratio 3.958 2.630 is very close to

1.5=32

. So we conclude that the gas has three moles

O to two moles N making the empirical formula

N2O3.

<h2><u>Mark as Brainliest</u></h2>
5 0
2 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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