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Mrrafil [7]
3 years ago
10

This image represents what?

Chemistry
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

nitrogen atoms

Explanation:

You might be interested in
If you started with 35.0 grams of H2S and 40.0 grams of O2, how many grams of S8 would be produced, assuming 95 % yield? 8H2S(g)
Vesnalui [34]
Molar mass

H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol

Identifying excess reagent and the limiting of the reaction :

8 H2S(g) + 4 O2(g) = S8(I) +  8 H2O(g)

8 x 34 g H2S -------->  256. 52 g S8
35.0 g ----------------> ??

35.0 x 256.52 / 8 x 34 =

8978.2 / 272 => 33.00 g of S8  

H2S is  the  limiting reactant
---------------------------------------------

4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??

40.0 x 256.52 / 4 x 31.99 =

10260.8 / 127.96 = 80.16 g of S8  

O2 is the excess reagent is the excess <span>reagent
</span>
------------------------------------------------------------

H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced 

33.0 g ----------- 100%
?? g ------------- 95 %

95 x 33.00 / 100 => 31.35 g

hope this helps!


4 0
3 years ago
A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions?0.0800 M Na2CO3 0
AysviL [449]

Answer:

  • <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>

Explanation:

Assume 100% ionization:

<em><u>1) 0.100 M solution K₂SO₄</u></em>

  • K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)

  • Mole ratios: 1 mol K₂SO4 : 2 mol K⁺  + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3

  • At constant volume, the mole ratios are equal to the concentration ratios (M).

  • 1  M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions

This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.

<u>2) 0.0800 M Na₂CO₃</u>

  • Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻

  • 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions

This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.

<u>3)  0.100 M NaCl </u>

  • NaCl → Na⁺ + Cl⁻

  • 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions

This is not equal to 0.300 M ion, so not a correct option.

<u>4) 0.0750 M Na₃PO₄</u>

  • Na₃PO₄ → 3Na⁺ + PO₄³⁻

  • 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions

Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.

<u>5)  0.0500 M NaOH </u>

  • NaOH → Na⁺ + OH⁻

  • 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions

Not equal to 0.300 M, so wrong choice.

7 0
3 years ago
Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) When 0.17 mo
BigorU [14]

Answer:

K remains the same;

Q < K;

The reaction must run in the forward direction to reestablish the equilibrium;

The concentration of PCl_3 will decrease.

Explanation:

In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.

Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.

The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.

As a result, since PCl_3 is also a reactant, its concentration will decrease.

8 0
3 years ago
Which of the following would represent a single displacement reaction between Potassium Bromide (K^+Br^-) and Iodine (I), which
ra1l [238]

Answer:

Option : KBr + I -> KBr+I

Explanation:

Single-replacement reaction or single displacement reactions are a type of chemical reactions in which a whole compound reacts with an element in such a way that the element takes place of one of the compound's own elements and sets it free.

If we talk about KBr and I displacement reaction is not possible among these because Iodine is less reactive than Bromine that is why it will not react with KBr or replace Br.

                       KBr + I -> KBr+ I


Potassium Bromide + Iodine -> Potassium bromide + Iodine


Hope it help!

5 0
3 years ago
A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1". KOH + Cu(NO3)2 → KNO
svet-max [94.6K]

2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2

2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺

2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻

5 0
3 years ago
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