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nata0808 [166]
3 years ago
5

5.(08.02 MC)

Chemistry
1 answer:
igomit [66]3 years ago
7 0

Answer: 1.09

Explanation:

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For the balanced chemical reaction
musickatia [10]

Answer:

150

Explanation:

  • C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

  • 100 molecules CO₂ * \frac{6moleculesO_2}{4moleculesCO_2} = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

5 0
3 years ago
A cook places a pan of food on a hot stove and leaves it there. When the food is just about done cooking, which of the following
rjkz [21]

Answer:

C.

The pan will be the same temperature as the stove.

Explanation:

7 0
3 years ago
as 100 milliliters of 0.10 molar KOH is added to 100 milliliters of 0.10 molar HCL at 298 K, the pH of the res?A. decrease to 3B
Mila [183]

Answer:

C. increase to 7.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

KOH+HCl\rightarrow KCl+H_2O

Thus, the molar relationship is 1 to 1, therefore, the moles are:

n_{HCl}=0.1mol/L*0.1L=0.01molHCl\\n_{KOH}=0.1mol/L*0.1L=0.01molKOH\\

Thus, since the entire hydrogen ions are neutralized, the pH C. increase to 7.

Best regards.

8 0
3 years ago
Determine the number of ions produced in the dissociation of the compound listed. AlF3
scoray [572]
Answer is: the number of ions produced in the dissociation of aluminium fluoride is 4.
<span>
Chemical dissociation of aluminium fluoride in water:
AlF</span>₃(aq) → Al³⁺(aq) + 3F⁻(aq).<span>
There are four ions, one aluminium cation and three fluoride anions.
Aluminium has oxidation +3, because it lost three electrons, to have electron configuration as noble gas neon and fluorine has oxidation -1, because it gain one electron to </span>have electron configuration as noble gas neon.
6 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
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