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2 years ago

Which mineral is used in batteries

1 answer:

4 0

Batteries come in all shapes and sizes, and depending on the type of battery, the minerals that compose them are mainly lithium, cobalt, graphite, nickel and manganese.

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Brainiest The Alpine Marmot is a kind of large ground squirrel that lives in the mountains and eats grasses, herbs, grains and s

Ghella [55]

It's the second one. Their body activity decreases, mostly because they're sleeping and they conserve the food in their bodies for long periods of time.
<span>the marmots' body activity decreases to conserve food stored in their bodies.</span>

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2 years ago

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If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed. A) saturated

dem82 [27]

The answer is A) saturated

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2 years ago

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How can you increase the energy of a glass of water?

balu736 [363]

B. Heat is a form of energy so boiling it would increase energy. (I guess)

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2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the

nirvana33 [79]

Answer:

The answer to your question is P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?

Process

- To solve this problem use the Combined gas law.

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

1000 ml -------------------- 1 l

855 ml -------------------- x

x = (855 x 1) / 1000

x = 0.855 l

-Substitution

P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

P2 = 377600 / 141.075

-Result

P2 = 2676.6 kPa

3 0

2 years ago