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emmasim [6.3K]
3 years ago
5

What is the location on the surface direction above the earthquake focus called

Chemistry
1 answer:
BARSIC [14]3 years ago
6 0
An earthquake's focus<span> is </span>located<span> along a fault beneath Earth's </span>surface<span>. The </span>focus<span> is the point within Earth where an </span>earthquake<span> starts. The </span>epicenter<span> is the </span>location on the surface<span> directly </span>above<span> the </span>focus<span>.</span>
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When preparing a buret for use in lab, the buret must always be cleaned thoroughly. prior to being filled with solution, the bur
erastova [34]
It is rinsed one last time with the solution to be measured because if there is water in the burret, then it could alter the results. Slightly, but it is still altering it.
6 0
3 years ago
Which of the following equations is balanced?
Alex_Xolod [135]
Where are the equations ..

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7 0
4 years ago
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How do you figure out molality
hodyreva [135]
*The molality of a solution is calculated by taking the moles of solute and dividing by
the kilograms of solvent* Basically if we had 1.00 mole of sucrose (it's about 342 3 grams) and
proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar
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What would be the molality of this solution? Notice that my one liter of water weighs
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3 0
2 years ago
What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>

4 0
3 years ago
Read 2 more answers
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