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anygoal [31]
2 years ago
7

In 1774, Joseph Priestly discovered oxygen gas by heating mercury(II) oxide powder with focused sunlight through a lens. How man

y grams of HgO (216.59 g/mol) would be decomposed to produce 1.50 mol O2 ? A. 650. g HgO B. 325gHgO C. 162gHgO D. 3.00 g HgO

Chemistry
1 answer:
mars1129 [50]2 years ago
3 0

Answer:

A. 650 g

Explanation:

The equation of the reaction is given as;

2 HgO(s) --> 2Hg(s)  + O2(l)

From the reaction;

2 mol of HgO produces 1 mol of O2

x mol of HgO would produce 1.50 mol of O2

2 = 1

x = 1.5

Solving for x;

x = 1.5 * 2 / 1 = 3 mol

Converting to mass;

Mass = Number of moles * Molar mass

Mass = 3 mol  * 216.59 g/mol

Mass = 649.77 g ≈ 650. g

Correct option is ;

A. 650 g

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I believe that the percussion instruments are the drum and cymbal.

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Is this a model of an element, a compound, or a mixture? Explain your reasoning.
grandymaker [24]

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A mixture

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1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?
Sav [38]
Wave speed = frequency x wavelength 
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Hope this helps!
5 0
3 years ago
Read 2 more answers
If you had a 0.5 M KCl solution, how much solute would you have in moles, and what would the solute be?
Svetradugi [14.3K]

Answer:

37.25 grams/L.

Explanation:

  • Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>

<em></em>

∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)

∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))

∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.

<em>So, the grams/L of KCl = 37.25 grams/L.</em>

6 0
3 years ago
Transamination of an amino acid transfers an amine group to form an α‑keto acid and is catalyzed by transaminases. Some amino ac
iragen [17]

Answer:

Amino acids that fall under the first category are alanine, aspartate and glutamate. While amino acids that fall under the second category are glycine, valine, proline, leucine, isoleucine, methionine, serine, threonine, cysteine, asparagine, glutamine, phenylalanine, tryptophan, tyrosine, lysine, arginine and histidine

Explanation:

Alanine, a 3-carbon amino acid reacts with α-ketoglutarate, a 5-carbon ketoacid, to produce pyruvate, a 3-carbon compound which is one of the glycolytic products in aerobic respiration, and glutamate, a 5-carbon amino acid, with the aid of alanine transaminase (ALT). The amino group from alanine is transfered from the α-carbon of alanine to the α-carbon of the α-ketoglutarate.

Aspartate, a 4-carbon amino acid also reacts with α-ketoglutarate to form oxaloacetate, a 4-carbon ketoacid which is present as an intermediate in the citric acid cycle, and glutamate. The amino group transfer occurs between the α-carbon of aspartate and the α-carbon of α-ketoglutarate.

Unlike the presence of a ketoacid in the conversion of alanine and aspartate to their corresponding amino acids, glutamate conversion to α-ketoglutarate, an intermediate in the citric acid cycle, involves no ketoacids as an amino group acceptor from glutamate. The amino group is freely released as an ammonium ion. The reaction involves the presence of a cofactor, NAD+ or NADP+, water (H2O) and glutamate dehygrogenase (GDH).

Other amino acids involve several metabolic steps to be converted into glycolytic or citric acid intermediates.

7 0
2 years ago
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