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allochka39001 [22]
3 years ago
15

I don’t know the answer

Chemistry
2 answers:
Y_Kistochka [10]3 years ago
5 0
D : chloroplasts , it’s a plant cell
Margarita [4]3 years ago
3 0
D. Chloroplasts bc they help the plant make foos
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what is the limiting reactant in the reaction of 12.0 g of SO2 with 8.0 g of H2S and their reaction? what mass of sulfur will be
larisa86 [58]
The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂ 
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂ 
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
7 0
3 years ago
If oxygen at 128 kpa is allowed to expand at constant temp until it's pressure is 101.3 kpa how much larger will the volume beco
LenaWriter [7]

which means that the volume increased by 26.4 mL in order to compensate for the decrease in pressure.

Like I said, depends on what your initial volume was, but that's how you think of it.

Hope this helped!

6 0
3 years ago
Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0
slavikrds [6]

Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$    -----     Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$   ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$   -----   Similar solubility as in the pure water

$0.10 \ M \ KCH_3COO$   -----   Similar solubility as in the pure water

The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of Ag^+, so the solubility of Ag_2CrO_4 decreases.

Both AgCH_3COO and $KCH_3COO$ are neutral mediums, so they do not affect the solubility.

 

4 0
3 years ago
The enthalpy of formation of MX is ΔHf° = –525 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 139 kJ/mol. The ionization en
polet [3.4K]
I'm sorry but could you tell me IE2 I couldn't asks for details
3 0
3 years ago
Why this process is called a cycle
Greeley [361]
<span>because it is a pattern; airgo cycle</span>
5 0
3 years ago
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