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Nataly [62]
3 years ago
14

How many molecules of sulfur dioxide are present in 1.60 moles of sulfur dioxide

Chemistry
1 answer:
ELEN [110]3 years ago
5 0
For this question, you must know that there are 6.022e23 atoms/molecules per mole of any substance (this is Avogadro's number). Therefore, your answer is 6.022e23 * 1.60 = 9.64e23 molecules of sulfur dioxide. (the "e" represents "times ten to the power of ___ ")
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Give an example of a mutation and explain how it could affect an organism?<br> Your answer
ella [17]

Answer:

A mutation is a heritable change in the genetic material of an individual. The change can be large or small. Large changes involve the loss, addition, duplication, or rearrangement of whole chromosomes or chromosome segments. Mutations can affect an organism by changing its physical characteristics (or phenotype) or it can impact the way DNA codes the genetic information (genotype). When mutations occur they can cause termination (death) of an organism or they can be partially lethal.

Explanation:

8 0
3 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
PolarNik [594]
V=3,5L\\&#10;Cm=0,9M\\&#10;M_{KCl}=74\frac{g}{mol}\\\\&#10;C_{m}=\frac{n}{V}\\\\&#10;n=\frac{m}{M}\\\\&#10;C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\&#10;m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
4 0
3 years ago
Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r
Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid,  the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄  NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).  

5 0
3 years ago
Someone help me i don’t understand it
Troyanec [42]
Use PV=nRT to help you
3 0
3 years ago
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
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