All categories

3 years ago

HELP ME PLEASE NEED HELP URGENTLY

Chemistry

1 answer:

kenny6666 [7]3 years ago

6 0

Answer: c

Explanation:

In case of 4 mols of product, we need 4/2 mole ratio x 130 = 260 Kcal

You might be interested in

As the energy level decreases, the size of the orbital:

hoa [83]

Answer:

B or C

Explanation:

When an atom absorbs the right amount of energy, an electron moves from its original orbital to a higher-energy orbital that has a vacancy. Similarly, when an atom emits energy, the electron drops to a lower-energy orbital that has a vacancy.

This is all I could find

6 0

3 years ago

Read 2 more answers

Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3

NARA [144]

Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

As it is given that $q_{1} = 1 \times 10^{-8} C$ , $q_{2} = 2 \times 10^{-8} C$ , and $q_{3} = 3 \times 10^{-8} C$ .

Distance between the charges = 1 cm = $1 \times 10^{-2} m$ (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

= $9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]$

= $9 \times 10^{9} [2 + 6 + 1.5]$

= $85.5 \times 10^{-5}$ J

= 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.

8 0

3 years ago

For the reaction, a → b, the rate constant is 0.0208 m-1 sec-1. How long would it take for [a] to decrease from 0.100 to 0.0450

goldfiish [28.3K]

First, assume the order of the given reaction is n, then the rate of reaction i.e. $\frac{dx}{dt}=k\times[A]^{n}$

where, dx is change in concentration of A in small time interval dt and k is rate constant.

According to units of rate constant, the reaction is of second order.

$\frac{1}{[A]_{t}}-\frac{1}{[A]_{o}} = kt$ (second order formula)

Put the values,

$\frac{1}{0.04590 m}-\frac{1}{0.100 m} =0.0208 m^{-1}s^{-1} \times t$

$22.23 m -10 m =0.0208 m^{-1}s^{-1} \times t$

$\frac{12.23 m}{0.0208 m^{-1}s^{-1}} = t$

t= 587.9 s

Hence, time taken is 587.9 s

5 0

3 years ago

What kind of bonds involve a transfer of electrons? what kind of bond involves sharing of electrons?

Ray Of Light [21]

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

3 0

3 years ago

Read 2 more answers

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.50 moles of magnesium perchlorate, mg(clo4)2. expres

Eva8 [605]

<span>4.50, 9.00, 36.00 First, count the number of atoms of each element in a single molecule of Mg(ClO4)2. Mg = 1 atom Cl = 1 * 2 = 2 atoms O = 4 * 2 = 8 atoms Now multiply the number of atoms of each element per molecule of the compound by the number of moles of the compound. So Mg = 1 * 4.50 = 4.50 Cl = 2 * 4.50 = 9.00 O = 8 * 4.50 = 36.00 So the answer is 4.50, 9.00, 36.00</span>

7 0

4 years ago