A karyotype can be used to determine the number of ____________________ (alleles, chromosomes) in a person’s cells.

A piece of iron (mass = 25.00 g) at 398.0 K is placed in a styrofoam coffee cup containing 25.00 mL of water at 298.0 K. Assumin

Cloud [144]

Answer:

The final temperature of the water is 36,93°C

Explanation:

When iron comes into contact with water, it gives it heat to increase its temperature. The heat that the iron yields to match the water temperature is.

Q = m . C . ΔT

Q = 25 g . 0,4990 J/g°C (298K - 398K)

Q = 25 g . 0,4990 J/g°C - 100°C = -1247,5 J

Kelvin or Celsius is the same, because ultimately the formula only takes the result of a subtraction

This is the heat <u>(-1247,5 J)</u> that water wins to increase its temperature. As in iron it is a detached heat, the result was negative, however, for water it is a gained heat, that's why it is positive.

1247,5 J = 25g . 4,180 J/g°C (Tfinal - 25°C)

For units, it's more confortable to work in Celsius.

1247,5 J / (25g . 4,180 J/g°C) = Tfinal - 25°C

11,93°C = Tfinal - 25°C

11,93°C + 25°C = Tfinal = 36,93 °C

It is reasonable that the water has increased its temperature because the iron temperature was 100 degrees higher, before both substances came into contact

5 0

3 years ago

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the followin

olya-2409 [2.1K]

Answer:

a) 0.70

b) 7.00

c) 0.85

d) 12.15

e) 1.30

Explanation:

The neutralization reaction involved in the titration is:

HClO₄(aq) + KOH(aq) → KClO₄(aq) + H₂O(l)

According to the chemical equation, 1 mol of HClO₄ reacts with 1 mol of KOH (1 equivalent of acid with 1 equivalent of base). The moles are calculated from the product of the molar concentration (M) and the volume in liters.

We have the following moles of acid (HClO₄):

40.0 mL x 1 L/1000 mL = 0.04 L

0.200 mol/L x 0.04 L = 8 x 10⁻³ moles HClO₄

Since HClO₄ is a strong acid (completely dissociated into H⁺ and ClO₄⁻ ions), the moles of HClO₄ are equal to the moles of H⁺. Then, we can calculate the initial pH:

[H⁺] = 0.200 M → pH = -log [H⁺] = -log (0.200) = 0.70

Now, we calculate the pH after the addition of KOH. Since KOH is a strong base, the concentration of KOH is equal to the concentration of OH⁻ ions.

a) 0.0 mL

No KOH is added, so the pH is the initial pH: 0.70

b) 80.0 mL KOH

80.0 mL x 1 L/1000 mL = 0.08 L

0.100 mol/L x 0.08 L = 8 x 10⁻³ moles KOH = 8 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 8 x 10⁻³ moles OH⁻ = 0

The neutralization reaction is complete and there is no remaining H⁺ from the acid. The concentration of H⁺ is equal to the concentration of H⁺ of water:

[H⁺] = 1 x 10⁻⁷ M → pH = -log [H⁺] = -log (1 x 10⁻⁷) = 7.0

c) 10.0 mL KOH

10.0 mL x 1 L/1000 mL = 0.01 L

0.100 mol/L x 0.01 L = 1 x 10⁻³ moles KOH = 1 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 1 x 10⁻³ moles OH⁻ = 7 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 10.0 mL = 50 mL = 0.05 L

[H⁺] = 7 x 10⁻³ moles/0.05 L = 0.14 → pH = -log [H⁺] = -log (0.14) = 0.85

d) 100.0 mL KOH

100.0 mL x 1 L/1000 mL = 0.1 L

0.100 mol/L x 0.1 L = 0.01 moles KOH = 1 x 10⁻² moles OH⁻

After neutralization we have:

1 x 10⁻² moles OH⁻ - 8 x 10⁻³ moles H⁺ = 2 x 10⁻³ moles OH⁻

The total volume is: V = 40.0 mL + 100.0 mL = 140 mL = 0.14 L

[OH⁻] = 2 x 10⁻³ moles/0.14 L = 0.014 → pOH = -log [OH⁻] = -log (0.014) = 1.84

pH + pOH = 14 → pH = 14 - pOH = 14 - 1.84 = 12.15

e) 40.0 mL KOH

40.0 mL x 1 L/1000 mL = 0.04 L

0.100 mol/L x 0.04 L = 4 x 10⁻³ moles KOH = 4 x 10⁻³ moles OH⁻

After neutralization we have:

8 x 10⁻³ moles H⁺ - 4 x 10⁻³ moles OH⁻ = 4 x 10⁻³ moles H⁺

The total volume is: V = 40.0 mL + 40.0 mL = 80.0 mL = 0.08 L

[OH⁻] = 4 x 10⁻³ moles/0.08 L = 0.05 M → pH = -log [H⁺] = -log (0.05) = 1.30

5 0

3 years ago

3. Svante Arrhenius developed a definition of acids and bases

Alexus [3.1K]

Answer:

Explanation:

BIG OL BEAN

6 0

3 years ago

A Questic

Ivanshal [37]

Answer:

No, it is not feasible because the Gibbs free energy change is positive

Explanation:

∆Hreaction= (-602 KJ/mol) - (-348 KJ/mol) = -254 KJ/mol

∆Sreaction = (42 + 27) J/Kmol - (33 + 44) J/K = -8J/Kmol

From;

∆G = ∆H - T∆S

∆G = 254 × 10^3 J/mol - [340K × (-8 J/Kmol)]

∆G = 2.57 × 10^5 J/mol

Note that when the change in free energy is positive, a reaction is non spontaneous. Only a reaction that has a negative change in free energy is spontaneous.

3 0

3 years ago

PLEASE HELP! The cooling curve below represents the cooling of a sample of gas. According to the cooling curve, what phase(s) wo

kiruha [24]

According to the cooling curve given above, at temperature 275K, the phases that will be observed is LIQUID AND SOLID.
The flat points on the graph represent the points where phase changes are taking place.The beginning of the first flat point represent the point where the gas become liquid, the extreme end of this point indicate the point where the liquid has started turning to liquid and that is the point of 275K. Thus at that point both phases are present.

5 0

4 years ago