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Olegator [25]
3 years ago
12

What are the primary components of rocks?

Chemistry
1 answer:
Butoxors [25]3 years ago
4 0
The primary components of rocks are minerals.
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If 7.400 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temp
Andru [333]
In your problems the possible solution is this one:

<span><span>F<span>fr</span></span>=<span>μk</span><span>FN</span>=<span>μk</span>mg</span><span>0.5(50000kg)(−9.8<span>m<span>s2</span></span>)</span><span>=−245000N</span> <span>a=F/m</span><span>−245000N/50000kg</span><span>=−4.9<span>m<span>s2</span></span></span><span>t=<span><span>v−<span>v0</span></span>a</span></span><span><span>0−10<span>ms</span></span><span>−4.9<span>m<span>s2</span></span></span></span><span><span>=2s

I hope you understand and satisfied with my answer</span></span>
8 0
3 years ago
Find the molar mass of H20. _______ grams per mole
Tpy6a [65]
It would be: 1 * 2 + 16 = 2 + 16 = 18

In short, Your Answer would be Option B

Hope this helps!
4 0
3 years ago
Which of these is a element: Salad, Sugar, Water, or Potassium
vovikov84 [41]

Potassium is a chemical element

Symbol: K

Atomic number: 19

Atomic mass: 39.0983 u ± 0.0001 u

Electron configuration: [Ar] 4s1

Melting point: 146.3°F (63.5°C)

3 0
3 years ago
Why is the equation 4Al+302&gt;2Al203 balanced
cestrela7 [59]

Answer:

It's coefficient to the front of each element that requires it.

Explanation:

It is coefficient to the front of each element or compound that requires it. Essentially you are multiplying the amount of atoms or compounds on one side to match the amount on the other side.

8 0
3 years ago
Nuclear decay occurs according to first-order kinetics. What is the half-life of europium-152 if a sample decays from 10.0 g to
djyliett [7]

Answer:

13.5 years

Explanation:

Initial Concentration [Ao] = 10g

Final Concentration [A] = 0.768g

Time t= 50 years

Half life t1/2 = ?

These quantities are related by the following equations;

ln[A] = ln[Ao] - kt ......(i)

t1/2 = ln(2) / k ...........(ii)

where k = rate constant

Inserting the values in eqn (i) and solving for k, we have;

ln(0.768) = ln(10) - k (50)

-0.2640 = 2.3026 - 50k

50k = 2.3026 + 0.2640

k = 2.5666 / 50 = 0.051332

Insert the value of k in eqn (ii);

t1/2 = ln(2) / k

t1/2 = 0.693 / 0.051332 = 13.5 years

4 0
3 years ago
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