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3 years ago

The resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is

Chemistry

1 answer:

Natalka [10]3 years ago

4 0

Answer: -

Thus the resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is the same as original.

Explanation: -

Let the mass number be A and the atomic number be Z.

When an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

Thus after alpha particle the new mass number = A - 4

Thus after alpha particle the new atomic number = Z - 2

When a positron is emitted, the atomic number only decreases by 1.

Thus after positron emission the new mass number = A - 4

Thus after positron emission the new atomic number = Z - 2 - 1

= Z - 3

When a beta particle is emitted, the atomic number increases by 1.

For 3 beta particles, the atomic number will increase by 3.

Thus after beta particle emission the new mass number = A - 4

Thus after beta particle emission the new atomic number = Z - 3+3

= Z

Thus the resulting atomic number of an element that emits 1 alpha particle, 1 positron, and 3 beta particles is the same as original.

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En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

Serga [27]

Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) ¿Cuál es la masa de agua a temperatura ambiente?

Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

$m_{o} =\rho_{w}\cdot V_{o}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{o} = 500\,mL$ , entonces:

$m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)$

$m_{o} = 500\,g$

La masa del agua a temperatura ambiente es de 500 gramos.

2) ¿Cuál es la masa de agua cuando se congela?

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) ¿Cuál es la masa de agua que queda después de la evaporación?

Durante la evaporación una parte del agua es transferida al aire, entonces podemos calcular la masa final ( $m_{f}$ ), medido en gramos, de la sustancia al multiplicar el volumen final ( $V_{f}$ ), medido en mililitros, por la densidad del agua ( $\rho_{w}$ ), medida en gramos por mililitro,. Es decir,

$m_{f} =\rho_{w}\cdot V_{f}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{f} = 400\,mL$ , entonces:

$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

$m_{f} = 400\,g$

La masa de agua que queda después de la evaporación es de 400 gramos.

4) ¿Qué masa de agua se evaporó?

Determinamos que la masa evaporada de agua ( $m_{v}$ ), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

$m_{v} = 500\,g -400\,g$

$m_{v} = 100\,g$

Se ha evaporado 100 gramos de agua.

5 0

3 years ago

Which of these lab safety procedures is INCORRECT?

Ksju [112]

Answer:

Explanation:

D is not necessary to follow lab safety in Laboratory

6 0

3 years ago

A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in

tamaranim1 [39]

Answer:

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Explanation:

Step 1: Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

Step 2: Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

Step 3: Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

Step 4: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

Step 5: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

5 0

3 years ago

The molar mass of gallium (Ga) is 69.72 g/mol.

Thepotemich [5.8K]

Answer:

2.35 x 10²⁰ atoms Ga

Explanation:

After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.

$27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga$

Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.

6 0

3 years ago

Group 2A metal oxides react with water to produce hydroxides. When 35.55 g of CaO are placed into 125 mL of water (d= 1.000 g/mL

Vika [28.1K]

Answer is: mass of calcium hydroxide is 46.98 grams.
Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.
m(CaO) = 35.55 g.
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 35.55 g ÷ 56 g/mol.
n(CaO) = 0.634 mol; limiting reactant.
m(H₂O) = 125 mL · 1.000 g/mL.
m(H₂O) = 125 g.
n(H₂O) = 125 g ÷ 18 g/mol.
n(H₂O) = 6.94 mol.
From chemical reaction: n(CaO) : n(Ca(OH)₂) = 1 : 1.
n(Ca(OH)₂) = 0.634 mol.
m(Ca(OH)₂) = 0.634 mol · 74.1 g/mol = 46.98 g.

5 0

3 years ago