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ExtremeBDS [4]
3 years ago
5

Fill in the blanks.

Chemistry
2 answers:
vichka [17]3 years ago
4 0

Answer:

Hello There!!

Explanation:

1.C nH 2n

2.Family

3.carbonyl group

4.Catalyst

5.Nickel

hope this helps,have a great day!!

~Pinky~

sweet [91]3 years ago
3 0

Answer:

\huge \purple {\mathfrak{Answer:}}

1. C nH 2n.

2. family .

3. carbonyl group.

4. catalyst.

5. Nickel.

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Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s
disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   HSO_4^-    ⇄  H^+ + SO_4^2^-

I    0.14

C   - x               +x       +x

E   0.14-x        x         x

K_a = 1.3 × 10^-^2 for HSO^-_4 . As a result,

\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a

K_a is large. It is no longer valid to approximate that [HSO^-_4] at equilibrium is the same as its initial value.

\frac{x^2}{0.14-y} = 1.3 * 10^-^2

x^2+1.3*10^-^2x - 0.14 × 1.3 × 10^-^2= 0

Solving the quadratic equation for x , x \geq 0 since x represents a concentration;

                             x=0.0366538

Then, round the results to 2 significant figure;

  • [SO_4^2^-] = x = 0.037 mol. L ^-^1
  • [H^+] = x = 0.037 mol. L ^-^1
  • [HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

3 0
2 years ago
Does a negative exponent mean that the number is less than 1 yes or no​
Masja [62]

Answer:

When a number is written in scientific notation, the exponent tells you if the term is a large or a small number. A positive exponent indicates a large number and a negative exponent indicates a small number that is between 0 and 1.

3 0
3 years ago
The product of the nuclear reaction in which 40Ar is subjected to neutron capture followed by alpha emission is ________. The pr
Orlov [11]

Answer:

37S

Explanation:

Radioactivity is the spontaneous emission of particles and / or electromagnetic radiation by unstable atomic nuclei leading to their disintegration.

We have two main types of radioactivity: radioactive decay and artificial transmutation.

In radioactive decay ( natural radioactivity ), a naturally occurring radioactive element like Uranium-238 disintegrates or decays into more stable isotopes with the emission of particles and/or radiation.

23892U = 23490Th + 42He

Artificial transmutation is the collision of two particles where one particle captures the other used to bombard it. There is subsequent production of isotopes similar or different from the bombarded particle. Neutrons, alpha particles ( helium nucleus ), electrons, protons can be used to bombard elements.

147N + 42He = 178O + 11P

For the above question which is artificial transmutation, the reaction equation is

4018Ar + 10n = 3716S + 42He

So, the neutron capture by Argon-40 will produce a radioisotope Sulphur-37 with the emission of an alpha particle.

5 0
3 years ago
What is the percent composition of the u.s quarter, which has the mass of 5.670g
Vitek1552 [10]
The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100

The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams

Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
6 0
3 years ago
in a typical person, the level of glucose is about 85 mg/ 100 mL of blood. if an average body contains about 11 pt of blood, how
Vanyuwa [196]

Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.

We make a proportion out of the word problem

(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose

5 0
3 years ago
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