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ohaa [14]
3 years ago
13

Once an neutralization reaction has occurred...

Chemistry
1 answer:
KatRina [158]3 years ago
6 0
I think the answer is b
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What most likely happens during this reaction
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I think that it is A I am sorry if I am wrong

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What Does M mean chemistry
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Molybdenum in periodic table

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Molarity definition

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Choose the correct answer from the choices below
marusya05 [52]

Answer:

either h or j

Explanation:

3 0
3 years ago
What is the limiting reactant in a reaction where 10.0 mol of iron is treated with 12.0 mol of bromine? The product that forms i
hammer [34]

<u>Answer:</u> The limiting reagent in the reaction is bromine.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

Given values:

Moles of iron = 10.0 moles

Moles of bromine = 12.0 moles

The chemical equation for the reaction of iron and bromine follows:

2Fe+3Br_2\rightarrow 2FeBr_3

By the stoichiometry of the reaction:

If 3 moles of bromine reacts with 2 moles of iron

So, 12.0 moles of bromine will react with = \frac{2}{3}\times 12.0=8moles of iron

As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Hence, bromine is considered a limiting reagent because it limits the formation of the product.

Thus, the limiting reagent in the reaction is bromine.

3 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
4 years ago
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