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3 years ago

15

Which formed first: hydrogen nuclei or hydrogen atoms?

1 answer:

Mama L [17]3 years ago

6 0

This sounds very much like a chicken-egg problem.

The first thing that formed must be hydrogen nuclei. The only other alternative is that the atom was created instantly, and the nuclei sprang forth at the same time as the atom, meaning that neither was technically first. The logic is that an atom can’t form without a nucleus, but it theoretically could be created instantly.

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When balancing a chemical equation, the number of h atoms in 2 ch4 is eight?

Viktor [21]

Yes... that is correct.

CH4 is methane so the coefficent in front of it would double the number of atoms of each element

6 0

3 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

3 years ago

True or false. Air masses mix at fronts?

Elina [12.6K]

The correct answer is False

6 0

3 years ago

what would the molarity of a solution be if you took 10 mL of a 13M stock solution and made a 300 mL solution?

nignag [31]

Answer:

0.43M

Explanation:

8 0

3 years ago

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33$

Therefore, the pH of a saturated solution is, 12.33

7 0

3 years ago