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2 years ago

What is not a product of modern chemistry ?

Chemistry

1 answer:

Doss [256]2 years ago

7 0

Answer:

Modern chemistry describes the composition, structure, and physical and chemical properties of substances. ... The main aim of modern chemistry is to improve the communication and conversation among the scientists, researchers, engineers, and policymakers, who are working under the area of modern chemistry

Explanation:

I hope this helps you!!

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You have been shipwrecked on a deserted island with no running water/fresh water, you have your clothes and a plastic bag contai

artcher [175]

Answer:

i have no clue

Explanation:

6 0

2 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

8 0

5 months ago

11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J

Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

∘

In your case, aluminium is said to have a specific heat of

0.90

∘

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

∘

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

∘

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

∘

, you'd have to provide it with

1 gram



0.90 J

1 gram



0.90 J

...

1 gram



0.90 J



10 times

0.90 J

However, you don't want to increase the temperature of the sample by

∘

, you want to increase it by

∘

−

∘

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

∘



0.90 J

∘



0.90 J

...

∘



0.90 J



33 times

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

∘

will be

10.0

⋅

0.90

∘

⋅

∘

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

⋅

, where

- the amount of heat added / removed

- the mass of the substance

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0

3 years ago

How many grams of CO are needed to react with an excess of Fe 2 O 3 to produce 209.7 g Fe?

gavmur [86]

Answer:

Mass = 157.5 g

Explanation:

Given data:

Mass of CO needed = ?

Mass of Fe formed = 209.7 g

Solution:

Chemical equation:

3CO + F₂O₃ → 2Fe + 3CO₂

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 209.7 g/ 55.85 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of iron and carbon monoxide.

Fe : CO

2 : 3

3.75 ; 3/2×3.75 = 5.625 mol

Mass of CO:

Mass = number of moles × molar mass

Mass = 5.625 mol × 28 g/mol

Mass = 157.5 g

4 0

2 years ago

If a “sample” of pennies contained 75 heads and 25 tails, how many half‐lives would have passed since the “sample” formed. Expla

finlep [7]

..............................

8 0

2 years ago