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3 years ago

13

PLZ ASAPThe model shows how the Sun fuses two nuclei of hellum-3 into beryllium-6. Since beryllium-6 is unstable, it will decay

into smaller nuclei. What additional product completes the model?

1 answer:

just olya [345]3 years ago

3 0

Answer:

Helium-4

Explanation:

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The overall reaction taking place is shown in the equation below

algol [13]

A. The maximum possible mass of sodium that can be produced is 47.18 Kg

B. The percentage yield of the reaction is 80.6%

<h3>Determination of the mass of NaCl and Na </h3>

We'll begin by calculating the mass of NaCl that reacted and the mass of Na obtained from the balanced equation.

2NaCl –> 2Na + Cl₂

Molar mass NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g = 117 / 1000 = 0.117 Kg

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g = 46 / 1000 = 0.046 Kg

SUMMARY

From the balanced equation above,

0.117 Kg of NaCl reacted to produce 0.046 Kg of Na.

<h3>A. How to determine mass of Na produced </h3>

From the balanced equation above,

0.117 Kg of NaCl reacted to produce 0.046 Kg of Na.

Therefore,

120 Kg of NaCl will react to produce = (120 × 0.046) / 0.117 = 47.18 Kg of Na

Thus, the maximum mass of Na produced is 47.18 Kg.

<h3>B. How to determine the percentage yield </h3>

Actual yield = 38.05 Kg
Theoretical yield = 47.18 Kg

Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (38.05 / 47.18) × 100

Percentage yield = 80.6%

Learn more about stoichiometry:

brainly.com/question/14735801

7 0

3 years ago

How much concentrated solution would it take to prepare 2.90 L of 0.420 M HCl upon dilution with water

Zanzabum

Answer:

the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)

Explanation:

the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂

moles of hydrochloric acid = M₁ * V₁= M₂ * V₂

V₁ = V₂ * M₂/M₁

where

M₂ = 0.420 M

V₂ =2.90 L

Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)

maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M

then

min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml

then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)

5 0

3 years ago

How many moles of gas does it take to occupy 25.6 liters at a pressure of 6.2 atmospheres and a temperature of 309.3 K?

Vilka [71]

Answer:

45 moles.

Explanation:

7 0

3 years ago

1) How many moles of gas occupy 58 L at a pressure of 2.2 atmospheres and a temperature

Sergio039 [100]

Here, we need to solve this problem using Ideal gas law ( PV=nRT).

Where –

P = Pressure in atm
V = Volume in L
n = moles
R = Ideal gas law constant
T = Temperature in K

<u>Now, according to the question </u>–

V = 58 L
P = 2.2 atm
T = 313 K
R = 0.0821 atm L/ mol K

<u>Calculation</u> –

$\qquad$ $\pink{\twoheadrightarrow\bf PV = nRT}$

$\qquad$ $\twoheadrightarrow\sf n = \dfrac{PV}{RT}$

$\qquad$ $\twoheadrightarrow\sf n = \dfrac{ 2.2 \: \:\times 58 \: } {0.0821 \times 313}$

$\qquad$ $\twoheadrightarrow\sf n = \dfrac{ 2.2 \times 58}{0.0821 \times 313}$

$\qquad$ $\twoheadrightarrow\sf n =\dfrac{127.6}{25.7}$

$\qquad$ $\pink{\twoheadrightarrow\bf n = 4.9656 \: moles }$

_______________________________________

8 0

3 years ago

Read 2 more answers

Lead metal is produced by heating solid lead (II) sulfide with solid lead (II) sulfate, resulting in liquid lead and sulfur diox

olya-2409 [2.1K]

Answer:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

Explanation:

The balanced reaction of heating solid lead (II) sulfide with the solid lead (II) sulfate to produce liquid lead and sulfur dioxide gas is shown below:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

In the balance reaction above, all the phases are indicated. This reaction is used for the production of lead metal.

4 0

3 years ago