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almond37 [142]
3 years ago
10

Stellar fusion produces atoms up to and including ___________ Question 51 options: Helium (atomic number 2) Carbon (atomic numbe

r 6) Iron (atomic number 26) Lead (atomic number 82) Question 52 (1 point)
Chemistry
1 answer:
Ganezh [65]3 years ago
3 0
<h2>Iron (Z = 26)</h2>

Explanation:

  • Stellar fusion is the nuclear reaction that occurs in stars.
  • In this reaction, the nuclei build to form heavier elements.
  • Stars fuse elements of lighter ones to heavier ones
  • This fusion reaction occurs in their cores.
  • During this process, a lot of energy is released.
  • This process is also known as stellar nucleosynthesis as there is a synthesis of the different nucleus.
  • Stellar fusion produces lighter elements of many types including iron(Fe) and nickel(Ni) in many of the giant stars.
  • These elements remain trapped in the cores of stars.
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Write the correct name for each compound. a. N2O5 b. Si2F4 c. S3F d. OF2 e. H4P6 f. C2O4 g. HF3
PSYCHO15rus [73]

Answer:

a. Boron trifluoride

b. Propane

c. Dinitrogen pentoxide

d. Carbon Dioxide

e. Silicon Octafluroride?

Explanation:

Glad to help :)

3 0
3 years ago
Hey do you guys have any ideas for a science fair project
Ierofanga [76]
You could do how different liquids affect plant growth. Take 3 SAME plant seeds and plant them in the SAME environment. For plant A use juice, for plant B, use milk, and for planting C, use coffee.
3 0
2 years ago
What is the variable that is changed by
denpristay [2]

Answer:

b. independent/manipulated variable

Explanation:

Independent/manipulated variable - refers to the variable that is changed by the scientist or an experimenter. Only one variable that is independent is required to ensure a fair test in an excellent experiment. As the independent variable is being changed by an experimenter or scientist, data is being recorded simultaneously as they are collected.

3 0
1 year ago
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0
3 years ago
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
Sedbober [7]

Answer:

Approximately 6.30\times 10^{-3}\;\rm mol.

Explanation:

The gallium here is likely to be produced from a \rm NaGaO_2\, (aq) solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral "\mathtt{(III)}" next to \rm Ga.  This numeral indicates that the oxidation state of the gallium in this solution is equal to +3. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of \rm Ga\, (s).

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium \mathtt{(III)} solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s.

Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C.

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}.

It takes three moles of electrons to deposit one mole of gallium atoms \rm Ga\, (s). As a result, \rm 1.89\times 10^{-2}\; mol of electrons would deposit \displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol of gallium atoms \rm Ga\, (s).

8 0
2 years ago
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