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3 years ago

The _________________ are either elements or compounds which are present prior to a chemical reaction.

Chemistry

2 answers:

Savatey [412]3 years ago

4 0

The missing word would be reactintes

mr Goodwill [35]3 years ago

3 0

Reactants would be the missing word in the sentence because they make up the products in a chemical reaction

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Balance the following KMnO4 + HCl = KCl + MnCl2 +H2O + Cl2

Vadim26 [7]

2KMno4+16hcl=2kcl+2mncl2+8h2o+5cl2

6 0

3 years ago

What is the Percent yield is 4.65 got copper is produced by 1.87 percent yield is got copper is produced when 1.87 grams of alum

konstantin123 [22]

Answer:

The percent yield is 70.486%

Explanation:

$3CuSO_{4}+2Al\longrightarrow Al_{2}(SO_{4})_{3}+3Cu\\\\Weight\:of\:Al=1.87g\\54g\:of\:Al\:will\:give\:190.5g\:of\:Cu\\$

$1.87g\:of\:Al\:gives=\frac{190.5\times1.87}{54}=6.597g\:of\:Cu\\ \\Weight\:of\:Cu\:obtained=4.65g\\Theoretical\:Yield=6.597g\\Experimental\:Yield=4.65g\\Percent\:Yield=\frac{Experimental\:Yield}{Theoretical\:Yield}\times100=\frac{4.65}{6.597}\times100=70.486\%$

Hence, the percent yield is 70.486%

7 0

3 years ago

Would magnesium prefer to lose or gain electrons

slega [8]

Umm what kind of question is that but i think gain lol

8 0

3 years ago

An atom of lithium 7 has an equal number of

Oliga [24]

Lithium 7 has an equal number of protons and neutrons.

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3 years ago

Read 2 more answers

For parts a & b below, derive only the initial value problem set up.

Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

$\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )$

The rate of the sugar coming into the tank is 0

$\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min$

$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So; y(t) = 25 $e^{-\dfrac{t}{100}$

Thus, the amount of sugar at any time t = $\mathbf{25 e^{^{-\dfrac{t}{100}}}}$

B) For tank B :

$\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}$

8 0

3 years ago