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Greeley [361]
3 years ago
12

I get it! Now I know that an organism's traits are controlled by ____________

Chemistry
1 answer:
Molodets [167]3 years ago
4 0

Answer:

Genes mainly, however the environment as well

Explanation:

You might be interested in
Which molecule, when added to a solution, would give the solution the highest pH
kondor19780726 [428]

Potassium Hydroxide (KOH) when added to the solution would give the highest pH

Explanation:

Bases or Alkali are associated with high pH while Acidic substances are represented by lower pH value.

In the given option

HCl- is a strong acid hence would have pH less than 7

H2SO4- also an acid with a pH less than 7

KOH- base with a pH higher than 7

H20-neutral compound with pH as 7

KOH is a very strong base and dissociates in aqueous solution to give it's corresponding metallic ion and hydroxyl ions (OH-) which are characteristic property of any base.

4 0
4 years ago
Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f
julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-

HNO_3 is Bronsted lowry acid and NO_3^- is its conjugate base.

H_2O is Bronsted lowry base and H_3O^+ is its conjugate acid.

(b)

CN^- + H_2O \rightarrow HCN + OH^-

CN^- is Bronsted lowry base and HCN is its conjugate acid.

H_2O is Bronsted lowry acid and OH^- is its conjugate base.

(c)

H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-

H_2SO_4 is Bronsted lowry acid and HSO_4^- is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O

HSO_4^- is Bronsted lowry acid and SO_4^{2-} is its conjugate base.

OH^- is Bronsted lowry base and H_2O is its conjugate acid.

(e)

O_{2-}+H_2O \rightarrow 2OH^-

O_{2-} is Bronsted lowry base and OH- is its conjugate acid.

H_2O is Bronsted lowry acid and OH- is its conjugate base.

6 0
4 years ago
I'd really appreciate help- this is due in around 8 hours :"D
Fed [463]

Answer:

  1. atomic #= 1
  • protons equals one
  • Number of neutrons equals to 0
  • number of electrons equal to 1
  • mass number equals to 1

2.

  • atomic number equals to 1
  • Protons equals to 1
  • number of neutrons equals to 0
  • number of electrons equals to 0 because the plus sign means that it loses an electron
  • mass number equals to 1

3.

  • atomic number 17
  • number of protons 17
  • number of neutrons 18
  • number for electrons 18 Because negative sign means It's an anion which means it gains an electron
  • mass number 35

4.

  • 12
  • 12
  • 12
  • 10 loses 2 its a cation
  • 24

5.

  • 47
  • 47
  • 108-47 = 61
  • 46 loses 1
  • 108

6.

  • 16
  • 16
  • 32-16 =16
  • 18 it gains 2 electrons
  • 32

there u go

7 0
3 years ago
What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised
musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 0.985 atm
  • Initial volume (V₁): 3.65 L
  • Final pressure (P₂): 861.0 mmHg
  • Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

7 0
3 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
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