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3 years ago

33.75 g of copper are reacted with 117.5 g silver nitrate. How much excess reactant would be unused or left over( in grams)?

Chemistry

1 answer:

Sophie [7]3 years ago

4 0

Why don’t you just look it up

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Which element is a halogen? argon bromine calcium lithium

almond37 [142]

It is (CI) bromine
because, Halogen element, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).

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3 years ago

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One source of aluminum metal is alumina, Al2O3. a. Determine the percentage composition of Al in alumina. b. How many pounds of

Gnom [1K]

Answer:

sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer

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3 years ago

Is density a physical property that can be used to identify matter

Ganezh [65]

Yes. density is a physical property that can be used to identify matter

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4 years ago

POINTS 50 - 30!!! in my bio to find some of my unanswered questions

alex41 [277]

Answer:

I have answered all 9/10 + this one so 10/10

Explanation:

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2 years ago

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 :

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

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3 years ago