The beginning expansion of the big bang happened relatively slowly compared to how it is expanding now.

What is the ph of the solution prepared by dissolving 0.927 mol hcl in enough water to make a 150 l aqueous solution?

Brilliant_brown [7]

Answer:

The answer to your question is pH = 2.2

Explanation:

Data

pH = ?

moles of HCl = 0.927

volume = 150 l

Process

1.- Calculate the Molar concentration of HCl

Molarity = moles / volume (L)

Molarity = 0.927 / 150

= 0.00618

2.- Calculate the pH

pH = -log [HCl]

Substitution

pH = -log [0.00618]

Result

pH = 2.2

8 0

3 years ago

A sequence of thermonuclear fusion processes inside massive stars can continue to transform the nuclei of elements such as carbo

nlexa [21]

Answer:

Iron is the element that is produced at the limit of the reaction.

Explanation:

In nuclear fusion 2 lighter nuclei are combined together into a single nucleus releasing a tremendous amount of energy in the process.

Up to fusion of iron the reaction of fusion is exothermic but when iron atom pops out the reaction becomes endothermic and it requires very high amount of energy to fuse iron atoms which is not available thus marking an end to the fusion reaction.

5 0

3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0

3 years ago

What four factors influence the rate of a homogeneous reaction.

Anika [276]

1) concentration or partial pressure of species
involved. 2) temperature • 3) presence of catalyst
4) nature of reactants.

5 0

2 years ago

How many significant figures does 0.0006510 of have?

arsen [322]

It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6

7 0

3 years ago

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