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vovangra [49]
3 years ago
6

For green plant cells, which are the substances that result from photosynthesis that are used in cellular respiration? A. oxygen

and water B. oxygen and glucose C. carbon dioxide and water D. carbon dioxide and glucose
Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

B. oxygen and glucose

Explanation:

Oxygen for respiration and glucose for energy utilization

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A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion
liberstina [14]
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
8 0
3 years ago
An electron is the largest part of an atom. True or False?
IrinaVladis [17]

Answer: False

Explanation:

Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom. Protons and neutrons are found in the nucleus. They group together in the center of the atom.

6 0
3 years ago
Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu
bagirrra123 [75]
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.

So the answer is:
8,12,48


</span>
3 0
3 years ago
Read 2 more answers
An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined
e-lub [12.9K]

Answer:

6.68 X 10^-11

Explanation:

From the second Ka, you can calculate pKa = -log (Ka2) = 6.187

The pH at the second equivalence point (8.181) will be the average of pKa2  and pKa3. So,

8.181 = (6.187 + pKa3) / 2

Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11

7 0
3 years ago
Read the following summary of the article
True [87]

Answer:

ea

Explanation:

zxds

6 0
3 years ago
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