Ask question

Ask question

All categories

faltersainse [42]

3 years ago

15

When volcanoes errupt, they can release oxides of sulphur that react with tiny droplets of water in the air to form sulphuric ac

id. Sulphuric acid can be considered a(n) __________.
A)
greenhouse gas
B)
primary pollutant
C)
biofuel
D)
secondary pollutant
E)
ecotone

1 answer:

leonid [27]3 years ago

8 0

Sulphuric acid is a secondary pollutant

You might be interested in

This chemical reaction is unbalanced: ___Pb(NO₃)₂(aq) + ___Li₂SO₄(aq) → ____PbSO₄(s) + ____LiNO₃(aq) Which coefficient should ap

kakasveta [241]

Answer: 2

Explanation:

4 0

4 years ago

Fluorine-21 has a half life of approximately 5 seconds. You have 340 grams, what fraction of the original nuclei would remain af

madreJ [45]

The half life equation is -->P(t) = Pi (0.5) ^ (t/c)
c is equal to the element to reach its half-life (5 seconds)t is equal to the duration of time the element is expose to (20 seconds)Pi is the initial amount (340)0.5 is the base of this exponential function to represent half-life.P(t) is the expression for the function of time

P(20) = 340 (0.5)^20/5P(20) = 340 (0.5) ^4P(20)= 21.25 grams
Fraction = P(t)/Pi = P(20)/Pi =21.25/340 =1/16
Therefore, when given 20 seconds, 340 grams of Fluorine-21 will degrade to 21.25 grams OR 1/16 of its original mass.
Hope this method helps! (This is my answer btw, I think you may have accidentally posted twice?)

7 0

4 years ago

Understanding how much of a product is produced in a reaction is referred to as ________ understanding the reaction?

777dan777 [17]

Hello!

Understanding how much of a product is produced in a reaction is referred to as Stoichiometrically understanding the reaction.

Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. The first to talk about stoichiometry was Jeremias Benjamin Ritcher, who said that "Stoichiometry is the science that measures the quantitative proportions or mass ratios of chemical elements that are involved in a chemical reaction".

To calculate how much of a product is produced in a reaction, Stoichiometry is used, applying the law of conservation of mass. That means that the amount of product can be calculated from the amounts of reactants if they are known.

7 0

3 years ago

Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!

e-lub [12.9K]

Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol

Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.

85.2 are in a 2 L solution that has a concentration of 0.6 mol/L

4 0

3 years ago

Read 2 more answers

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