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3 years ago

HELP ASAP PLEASE CHEMISTRY

Chemistry

2 answers:

Dimas [21]3 years ago

6 0

Answer:

The rate would be higher, and the concentration of products would be lower.

Explanation:

lubasha [3.4K]3 years ago

4 0

The correct answer is The rate would be lower, and the concentration of products would be higher.

The rate of reaction refers to the speed with which reactants are converted into products.

The rate of reaction decreases as the reaction progresses because more reactants are being converted into products.

As such, the reactant concentration decreases and the concentration of products increases.

Hence, after three seconds, the rate would be lower, and the concentration of products would be higher.

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Arrange the steps typically followed in the data preparation process in the order of occurrence. (Place the first step at the to

Pavel [41]

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3 0

2 years ago

Help me please im in 4th grade and online is frustrating

Snowcat [4.5K]

Answer:

Explanation:

The answer is A) because the A container has tightly packed molecules but not as tight as C so that means A is liquid, and C is a solid which makes B a gas

7 0

4 years ago

Give 3 example<br>of a<br>solid that sublimes.

tino4ka555 [31]

Here are a few examples :)

iodine (I2)

naphthalene

aresenic (As)

ferrocene

water (H2O)

carbon dioxide (CO2)

Hope this helps :)

5 0

3 years ago

150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o

N76 [4]

Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

Therefore, pH of the final solution = 4.22

3 0

4 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