Onsider this representation of an original DNA sequence: THE TALL MAN WAS SAD.

Chemistry

2 answers:

kenny6666 [7]4 years ago

7 0

Answer:

Substitution

Explanation:

just did it on edge

frutty [35]4 years ago

6 0

Answer:

Substitution

Explanation:

Just answered the question

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If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according

Svetradugi [14.3K]

Answer:

9.94 mL, the volume of ethanol needed

Explanation:

The reaction is:

C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

We convert the mass of the formed product to moles:

15 g . 1mol / 44g = 0.341 moles

2 moles of dioxide are produced by 1 mol of ethanol, in order to stoichiometry.

Therefore, 0.341 moles of CO₂ must be produced by (0.341. 1) / 2 = 0.1705 moles of alcohol.

We convert the moles to mass, and then, the mass to volume by the use of density.

0.1705 mol . 46 g / 1 mol = 7.84 g of ethanol

Ethanol density = Ethanol mass /Ethanol volume

Ethanol volume = Ethanol mass /Ethanol density → 7.84 g /0.789 g/mL =

9.94 mL

7 0

3 years ago

Read 2 more answers

copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

a_sh-v [17]

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

$KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it: