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2 years ago

How is a beta particle formed?

Chemistry

1 answer:

zalisa [80]2 years ago

6 0

Answer:

forms when a neutron changes into a proton and a high-energy electron .

hope this helps <3

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Thermochemistry

max2010maxim [7]

THe Answer is 105 J

hope this helps and please brainliest, I was first :)

4 0

3 years ago

A solution of sulfuric acid has a concentration of 0.0980 g/L. If the density of the acid is 1.84 g/mL, what is the concentratio

mariarad [96]

The answer is 98ppm.

The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.

6 0

4 years ago

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Lab activity - Law of Conservation of Mass with vinegar and baking soda.

lesya [120]

Answer: I BIG DUMB

Explanation:

7 0

3 years ago

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago

El pH de una solución acuosa 10^-14 M de ácido acético, a 25°C, es igual a, teniendo en cuenta que Ke = 1.76x10^-5

frez [133]

Answer:

$pH=14$

Explanation:

Hola!

En este caso, consideramos que la disociación de ácido acético ocurre:

$CH_3COOH\rightarrow CH_3COO^-+H^+$

Así, mediante la solución del equilibrio ácido, podemos calcular la concentración de iones hidronio que posteriormente sirven para calcular el pH de la solución, por tal razón, debemos calcular el equilibrio dada la constante de equilibrio y por medio de la ley de acción de masas en términos del cambio $x$ como cualquier problema de equilibrio:

$Ke=\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\\\\1.76x10^{-5}=\frac{x*x}{1x10^{-14}M-x}$

Resolviendo para $x$ , tenemos $x=0.999x10^{-14}$

Así, la concentración de hidrógeno es igual a x, por lo que el pH:

$pH=-log([H^+])=-log(0.999x10^{-14})\\\\pH=14$

Dicho valor tiene sentido desde que la concentración de hidrógeno es casi despreciable, por lo que se puede asumir que tiende a ser básica.

Saludos!

5 0

3 years ago

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