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wolverine [178]
3 years ago
5

I will give 25 points and make them brainlest if they solve this.

Chemistry
2 answers:
sdas [7]3 years ago
7 0

Answer:

B

Explanation:

serious [3.7K]3 years ago
6 0

Answer:

B - Carbon atoms found in current living things are the same carbon atoms that have always been a part of living things, they are just recycled.

You might be interested in
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
A mixture of 117.9 g of P and 121.8 g of O, reacts completely to form P4O6 and P4O10. Find the masses of P4O6 and P4O10
Alex17521 [72]

Answer:

Mass of P4O6=103.4

            P4O10=133.48

Explanation:

Balanced reaction is:

8P +8O_{2}  ⇒P_{4} O_{6} +P_{4} O_{10}

Both reactant completely vanishes as equivalent of bot are equal.

Moles of P=\frac{117.9}{31} =3.80

Moles of O_{2} =\frac{117.9}{31} =3.80

No. of moles of formed product are equal and is \frac{1}{8}th of mole of any of reactant.

Thus weight of  P_{4} O_{6} =\frac{3.80}{8}×220 ≈103.41

        weight of  P_{4} O_{10} =\frac{3.80}{8}×284 ≈133.48

4 0
3 years ago
Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
stiv31 [10]
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: C_5H_{10}S_2
4 0
4 years ago
Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?
Bogdan [553]

The second option only.

  • LiOH, Li₂SO₄.
<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

  • Ca(OH)₂ and
  • LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}.

The formula for calcium sulfate \text{CaSO}_4 in option A is spelled incorrectly. Why? The charge on each calcium \text{Ca}^{2+} is +2. The charge on each sulfate ion {\text{SO}_4}^{2-} is -2. Unlike \text{Li}^{+} ions, it takes only one \text{Ca}^{2+} ion to balance the charge on each {\text{SO}_4}^{2-} ion. As a result, \text{Ca}^{2+} and {\text{SO}_4}^{2-} ions in calcium sulfate exist on a 1:1 ratio.

2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}.

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}.

The ion {\text{NH}_4}^{+} acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}.

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0
3 years ago
How much heat is added if .0948g of water is increased in temperature by .728 degrees C?
Verdich [7]

Answer:

0.289J of heat are added

Explanation:

We can relate the change in heat of a substance with its increasing in temperature using the equation:

q = m*ΔT*S

<em>Where Q is change in heat</em>

<em>m is mass of substance (In this case, 0.0948g of water)</em>

<em>ΔT = 0.728°C</em>

<em>S is specific heat (For water, 4.184J/g°C)</em>

Replacing:

q = 0.0948g*0.728°C*4.184J/g°C

q = 0.289J of heat are added

5 0
3 years ago
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