As water evaporates from a solution, minerals become more concentrated and begin to form a rock. what type of rock is it.

Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o

Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M$

where,

$n_p$ = number of protons

$m_p$ = mass of one proton

$n_n$ = number of neutrons

$m_n$ = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is $_{15}^{31}\textrm{P}$

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

$\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399$

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0

3 years ago

How wouldo I do research on which cup would keep a hot drink warmer

marshall27 [118]

ccording to Michigan State University, heat is created when molecules in the liquid move in different directions and bang into one another. These fast moving particles hit the side of the container where they are located. Heat conduction causes the heat from the liquid to be transferred to the container. The container gets hotter while the liquid gets colder. The liquid also loses heat as the surface area is exposed to air. The air gets heated while the container and the cup cool down.

A thermos container keeps liquids hot because the tight lid prevents heat from escaping the container. The core of the thermos is also filled with insulation, which does not conduct heat as well, so the liquid inside the cup does not cool down as quickly. Most thermos containers also feature reflective exteriors that limit the heat lost to radiation. A Styrofoam cup is made up of 95 percent air. This air conducts heat, which draws the warmth from the liquid into the cu

4 0

4 years ago

H2 burning in O2 to form H2O (l) is an example of a system where the entropy of the universe decreases.

leva [86]

Answer:

B. False

Explanation:

the reaction is

H₂0 + O₂ = H₂O

as addition of oxygen means burning in the system which means heating or energy is necessary for the reactants. SO the reaction is spontaneous and exothermic .

As according to second law of thermodynamics

the change in enthalpy is greater than or equal to zero when the reaction is spontaneous, so enthalpy will increase in the above reaction.

H2 burning in O2 to form H2O (l) is an example of a system where the entropy of the universe increases.

6 0

3 years ago

A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 x 10−5 as Ka for ac

Anika [276]

Answer:

a. pH = 5.22

b. Acidic.

c. pH = 5.14

Explanation:

a. It is possible to find the pH of a buffer using Henderson-Hasselbalch equation (H-H equation):

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka (For acetic acid = 4.74), [A⁻] is molar concentration of conjugate base (Acetate salt) and [HA] concentration of the weak acid (Acetic acid).</em>

Replacing:

pH = 4.74 + log₁₀ [0.600M] / [0.200M]

<em>You use the concentration of the acetic acid and sodium acetate because you're adding equal volumes, doing the ratio of the species the same</em>

<em />

b. As the solution has a pH lower that 7.0, it is considered as a <em>acidic solution.</em>

c. When you add HCl to the buffer, the reaction is:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

<em>Where acetate ion reacts with the acid producing acetic acid.</em>

As you have 0.200L of the buffer, 0.100L are of the acetate ion and 0.100L of the acetic acid. Initial moles of both compounds and moles of HCl added are:

CH₃COO⁻: 0.100L ₓ (0.600mol / L) = 0.0600 moles

CH₃COOH: 0.100L ₓ (0.200mol / L) = 0.0200 moles

HCl: 3.0mL = 3x10⁻³L ₓ (0.034mol / L) = 0.00010 moles HCl

The moles added of HCl are the same moles you're consuming of acetate ion and producing of acetic acid. Thus, moles after the reaction are:

CH₃COO⁻: 0.0600 moles - 0.0001 moles = 0.0509 moles

CH₃COOH: 0.0200 moles + 0.0001 moles = 0.0201 moles

Replacing in H-H equation:

pH = 4.74 + log₁₀ [0.0509moles] / [0.0201moles]

<em />

8 0

3 years ago

Given that the vapor pressure of pure n-hexane and pure n-heptane at 25°c are 151.4 mmhg and 45.62 mmhg respectively, calculate

tatyana61 [14]

Answer is: total pressure is 108.96 mmHg.

p(n-hexane) = 151.4 mmHg; pressure of n-hexane.
p(n-heptane) = 45.62 mmHg.
χ(n-hexane) = 0.600; mole fraction.
χ(n-heptane) = 1 - 0.6 = 0.400.
Using Raoult's Law: <span>
p(total) = p(n-hexane) · χ(n-hexane) + p(n-heptane) · χ(n-heptane).
</span>p(total) = 151.4 mmHg · 0.6 + 45.32 mmHg · 0.4.
p(total) = 108.96 mmHg.

6 0

4 years ago