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3 years ago

12

One of the hopes for solving the world's energy problem is

1 answer:

lions [1.4K]3 years ago

6 0

Answer:

2.803013439419911 × 10⁻¹² J

Explanation:

Mass defect = mass of reactant - mass of product

(2.0140 + 3.01605) - (4.002603 + 1.008665)

5.03005 - 5.011268 = 0.018782 amu

mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg

mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg

E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s

E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J

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Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP

Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

4 0

2 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

3 years ago

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

Irina18 [472]

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

All you have to do is to create a ratio between the molecule and the oxygen atom.

5 moles of Mg3(PO4)2 (4x2 moles O/1 mole Mg3(PO4)2) = 40 moles of oxygen

5 0

3 years ago

What is the molarity of a solution that contains 1.1 moles of lithium in 0.5 liters of solution?

cluponka [151]

0.5 litres contain 1.1 moles
therefore 1 litre will have= 1*1.1/0.5
molarity=2.2M

8 0

3 years ago

Read 2 more answers

The stability of atomic nuclei is related to the

Gekata [30.6K]

Ratio of neutrons & protons .

6 0

3 years ago