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daser333 [38]
3 years ago
6

What happens to the matter in the body of an animal after it dies

Chemistry
1 answer:
sladkih [1.3K]3 years ago
4 0
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
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Define An Atom.<br>Ty!!!​
astra-53 [7]

Answer:

the basic unit of a chemical element.

Explanation:

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Describe a way to separate a mixture of sand and salt .
worty [1.4K]
Put water in it and the salt will dissolve and the sand will stay at the bottom
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A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c
topjm [15]
<span>C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
3 0
3 years ago
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A 1300 mL sample of gas with a molar mass of 71.0 g/mol at STP has what density?
Blizzard [7]

Answer:

0.055g/mL

Explanation:

Data obtained from the question include:

Molar Mass of the gass sample = 71g/mol

Volume of the gas sample = 1300 mL

Density =?

The density of a substance is simply mass per unit volume. It is represented mathematically as:

Density = Mass /volume.

With the above equation, we can easily obtain the density of sample of gas as illustrated below:

Density = 71g / 1300 mL

Density = 0.055g/mL

Therefore, the density of the gas sample is 0.055g/mL

6 0
3 years ago
Is anyone good at chemistry if so can someone help me please ?<br><br> (NO LINKS)
Mkey [24]

This requires familiarity with the different theories (or concepts) of acids and bases.

On the Arrhenius concept, an acid is a substance that produces an H⁺ ion in water such that the H⁺ concentration increases, and a base is a substance that produces an OH⁻ ion in water such that the OH⁻ concentration increases.

On the Brønsted–Lowry concept, an acid is a substance that donates a proton (which is basically an H⁺ ion) in a solvent, and a base is a substance that accepts a proton in a solvent.

On the Lewis concept, an acid is a substance that accepts an electron pair in a solvent, and a base is a substance that donates an electron pair in a solvent.

The concepts become progressively broader, i.e., the Arrhenius concept is the most restrictive and the Lewis concept is the least restrictive. As a corollary, an Arrhenius acid or base is also both a Brønsted–Lowry acid or base and a Lewis acid or base, respectively; a Brønsted–Lowry acid or base is not necessarily an Arrhenius acid or base, but an Arrhenius acid or base is also a Lewis acid or base, respectively. And finally, a Lewis acid or base may not necessarily be either an Arrhenius or a Brønsted–Lowry acid or base.

So, with the above concepts in mind, we can match the statements in column A with the type of acid or base in column B:

\begin{center}\begin{tabular}{ c c } 1 & Bronsted Lowry acid \\  2 & Bronsted Lowry base \\   3 & Arrhenius acid \\ 4 & Arrhenius base \\ 5 & Lewis base \\ 6 & Lewis acid\end{tabular}\end{center}

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