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Firlakuza [10]
3 years ago
10

The specific heat of aluminum is 0.900 J/gº C. A 400.0 g sample of aluminum at 20.0º C absorbs 2520. J of heat energy. What is t

he new temperature of the aluminum?
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0
I'm pretty sure it's 8 xd
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Water is composed of one atom of oxygen and two atoms of
egoroff_w [7]

Answer:

<em>Hydrogen.</em>

Explanation:

You've probably seen "H_{2}0" which is the formula for water. It means that there's 2 hydrogen atoms, and one oxygen atom, in one molecule of water.

<em>Hope this helps! Feel free to mark me Brainliest if you feel this helped. :)</em>

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3 years ago
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What is the average of atomic mass of carbon if 98.90% of the atoms are c-12 and 1.100% are c-13 atoms?
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2 years ago
When hydrogen peroxide (H2O2) is added to potassium iodide (KI) solution, the hydrogen peroxide decomposes into water (H2O) and
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D. Catalyst

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Because a Catalyst is a substance that causes or accelerates a chemical reaction without itself being affected.

5 0
2 years ago
How many protons, neutrons, and electrons does sodiums neutral atom have? Plz help.
rusak2 [61]

Answer:

So for your question, the Periodic Table tells us that sodium has an Atomic Number of 11, so there are 11 protons and 11 electrons. The Periodic Table tells us that sodium has an Atomic Mass of ≈23. So there are 23 - 11 = 12 neutrons.

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3 0
3 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
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