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3 years ago

Which of the following is the correct scientific notation for 0.0000782 g?

Chemistry

2 answers:

EleoNora [17]3 years ago

8 0

Answer: The correct scientific notation is 7.82 x 105 g

Explanation:

Dominik [7]3 years ago

7 0

The correct scientific notation is 7.82 x 105 g

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You are making lemonade and the calls for 6 cups Lemon juice (L), 3 Cups of sugar (S) and 5 cups of water (W) to make 12 Cups of

vova2212 [387]

Answer:

18 Cups

Submitted the assignment

Explanation:

4 0

2 years ago

Rank these acids according to their expected pKa values.

givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO- + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

3 0

3 years ago

The reaction that occurs in a Breathalyzer, a device used to determine the alcohol level in a person's bloodstream, is given bel

Umnica [9.8K]

Answer:

The rate of disappearance of C₂H₆O = 2.46 mol/min

Explanation:

The equation of the reaction is given below:

2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O

From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.

The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.

Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃

Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O

Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min

4 0

3 years ago

Which of the following metals may tarnish to a greenish-blue color called verdigris

hoa [83]

Copper. thats the way copper oxidized. like the statue of liberty

3 0

3 years ago

Read 2 more answers

On the basis of the information above, a buffer with a pH = 9 can best be made by using

telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for $\mathbf{H_3PO_4 , H_2PO^{-}_4 , HPO_4^{2-}}$ are $\mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \ 5\times 10^{-13}}$ respectively.

$\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}$

$\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}$

$\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}$

The reason while option D is the best answer is that, the value of pKa for both

$\mathbf{H_2PO^{-}_4 ,\ \& \ HPO_4^{2-}}$ lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

$\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}$

3 0

3 years ago