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taurus [48]
3 years ago
15

Aluminum bronze contains 92.0% copper and 8.0% aluminum. What maximum mass of aluminum bronze can be prepared from 73.5g of copp

er and 42.2g of aluminum? Initially I tried dividing 73.5g by 0.92, and I thought I have found the max mass of aluminum bronze. But then I tried using 70.0g of copper which gave me a result of about 82.17g of total mass of aluminum bronze. How should I calculate it accurately so that I can maximize the use of both metals?
Chemistry
1 answer:
Keith_Richards [23]3 years ago
6 0
The ratio of aluminum bronze components is:

92.0 Cu / 8.0 Al

The ratio of Cu and Al avilable is: 73.5 Cu / 42.2 Al

Then, it is evident that Al is in excess and Cu is the limitant material.

So, you need two work with the open proportion 92.0/ 8.0 = 73.5Cu / x

=> x = 73.5 * 8 / 92 = 6.39

Then, you can use 6.4 grams of Al and 73.5 grams of Cu to prepare 6.39g + 73.5g =  79.89.grams of Bronze.

I hope this helps.





 
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Answer:

C

Explanation:

If you add enough heat to a solid it eventually becomes a liquid

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Explain how glucose and starch are related
weeeeeb [17]

Answer:

Starch is the stored form of sugars in plants and is made up of a mixture of amylose and amylopectin (both polymers of glucose). ... The cells can then absorb the glucose. Starch is made up of glucose monomers that are joined by α 1-4 or α 1-6 glycosidic bonds.

Explanation:hope that helps you lots

4 0
3 years ago
A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?
kotykmax [81]
First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass

mass = 252.4795g however the 6.5L has only 2 sig figs so i would say

mass CaCl2 = 2.5 * 10 ^2 g
5 0
3 years ago
Read 2 more answers
After a reaction, a new compound contains 0.73 g Mg and 0.28 g N. What is the empirical formula of this compound?
FromTheMoon [43]

Answer:

Mg₃N₂

Explanation:

The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:

0,73g × (1mol / 24,305g) = 0,03 moles of Mg

0,28g × (1mol / 14,006g) = 0,02 moles of N

Dividing each value in 0,01 to obtain natural numbers:

0,03 moles of Mg / 0,01 = 3

0,02 moles of N / 0,01 = 2.

Thus, empirical formula is: <em>Mg₃N₂</em>

<em></em>

I hope it helps!

5 0
3 years ago
What is the equilibrium constant expression for the reaction below?<br> H2 (g) + I2 (g) → 2 HI(g)
brilliants [131]
<h3>Further explanation</h3>

Given

Reaction

H₂ (g) + I₂ (g) → 2HI(g)

Required

The equilibrium constant

Solution

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant for reaction  

pA + qB ⇒ mC + nD  

\large {\boxed {\bold {K ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

So for the above reaction :

\tt K=\dfrac{[HI]^2}{[H_2][I_2]}

8 0
3 years ago
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