The half-life of argon-39 is 269 years. It decays into krypton-39. After 1,076 years, what fraction of the original amount of argon-
1/16
1/4
1/2
1/8
Answer:
The correct answer is - 1/16.
Explanation:
The half-life is the time that is required to decay nuclei to the half amount of its original amount in a radioactive sample. It is represented by the t1/2. The time that is required to reduce to half of its initial value is known as the half-life of a particular sample.
Half-life of argon-39 = 269 years
total time is taken in complete decay = 1076 years
Number of half-lives required = 1076/269
= 4
So the original amount of the sample = (1/2)^n
n = number of half-life
= 1/2^4
= 1/16
The correct answer is = 1/16
The air temperature in a sealed, insulated box.. 20°C An ice cube at O'C is
placed in the box where it slowly melts. How and why does the melting of the ice
affect the air temperature in the box?
Melting is endothermic so the air temperature increases
Melting is exothermic, so the air temperature decreases
Melting is endothermic, so the air temperature decreases
Melting is exothermic, so the air temperature increases
Answer:
Melting is endothermic, so the air temperature decreases.
Explanation:
Hello there!
In this case, according to the given statement, it is possible to infer that the ice is melt because energy is applied to the ice and thereafter its temperature increases; this is possible because the joined particles of a solid substance need energy to undergo such a separation that they become more far away to each other and therefore transcend to the liquid phase due to the new molecules arrangement. Thus, the answer is Melting is endothermic, so the air temperature decreases because as the ice heats up, the air cools down as it gives it energy to the ice.
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BIOCR
1 Co
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Answer the question below. Use the rubric in the materials for help if needed.
>
Describe the process used to determine how many atoms of each element are in 2Ca3(PO4)2. Show all
your work.
Answer:
Six atoms of calcium, four of phosphorous and sixteen of oxygen for a total of twenty six
Explanation:
Hello there!
In this case, according to the molecular formula of the two moles of calcium phosphate:
[tex]2Ca_3(PO_4)_2[/tex]
Thus, in order to calculate the atoms of each atom, it is necessary to multiply the two in front of the formula by the subscripts in the reaction:
[tex]atoms Ca=2*3=6\\\\atoms P=2*2=4\\\\atoms O=2*2*4=16[/tex]
Thus, we obtain six atoms of calcium, four of phosphorous and sixteen of oxygen for a total of twenty six.
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Is a cold and b is warm? Please help
Answer:
Label A shows cold area on the mug, while label B shows a warm area.
Explanation:
Have a good day
Is lighting a match a chemical change?
Answer:
yes
Explanation:
Lighting a match and letting is burn is an example of a chemical change.
Matches use sulfur, phosphate and a friction agent held together by a binding agent. Together, the oxygen and sulfur burn slowly, igniting the wood of the match for a flame that lasts long enough to see by, light a candle or ignite a campfire.
Which compound contains three elements?
A. Aluminium chloride
B. iron(III) oxide
C. potassium oxide
D. sodium carbonate
Answer:
Sodium Carbonate
Explanation:
Sodium carbonate is made of three elements, that are, sodium, carbon and oxygen.
Its formula is Na2CO3.
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Consider the reaction, C2H4(g) + H2(g) - C2H6(8), where AH = -137 kJ. How many kilojoules are released when 3.5 mol of CH4
reacts?
480 kJ are released
20 x 103 kJ are released
570 kJ are released
137 kJ are released
Answer: 480 kJ of energy is released when 3.5 mol of [tex]C_2H_4[/tex] reacts.
Explanation:
The balanced chemical reaction is:
[tex]C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)[/tex] [tex]\Delta H=-137kJ[/tex]
Thus it is given that the reaction is exothermic (heat energy is released) as enthalpy change for the reaction is negative.
1 mole of [tex]C_2H_4[/tex] on reacting gives = 137 kJ of energy
Thus 3.5 moles of [tex]C_2H_4[/tex] on reacting gives = [tex]\frac{137}{1}\times 3.5=480 kJ[/tex] of energy
Thus 480 kJ of energy is released when 3.5 mol of [tex]C_2H_4[/tex] reacts.
1.20×10−8s to nanoseconds
Answer:
There are 12 nanoseconds in [tex]1.2\times 10^{-8}\ s[/tex].
Explanation:
We need to convert [tex]1.2\times 10^{-8}\ s[/tex] to nanoseconds.
We know that,
[tex]1\ s=10^9\ ns[/tex]
Now using unitary method to solve it such that,
[tex]1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns[/tex]
So, there are 12 nanoseconds in [tex]1.2\times 10^{-8}\ s[/tex].
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What kind of weather forms with an occluded front?
Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.
DATA
Copper
Lead
Mass of Metal
Initial Temperature of Metal
Mass of Water
Initial Temperature of Water
Final Temperature of Water
CALCULATIONS
Copper
Lead
Temperature Change of Water
Heat Gained by Water
Heat Lost by Metal
Temperature Change of Metal
Specific Heat of Metal
% Error
Use the Gizmo to mix 200 g of lead at 100 °C with 1,000 g of water at 20 °C. Record the data and answers to the calculations for lead in the 2 tables above.
Metal
Accepted Specific Heat (J/g⁰C)
aluminum
0.900
lead
0.160
copper
0.385
tin
0.228
steel
0.460
Insert a picture of your work for the calculations table here.
Answer:
Explanation:Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.
DATA
Copper
Lead
Mass of Metal
What causes the lines in the spectrum for elements
How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?
Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
Amount of heat required to vaporize 1 mole of lead = 177.7 kJ
Molar mass of lead = 207.2 g
Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)
Heat required to vaporize 207.2 of lead = 177.7 kJ
Thus Heat required to vaporize 1310 g of lead =[tex]\frac{177.7}{207.2}\times 1310=1123kJ=1123000J[/tex]
Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
What is the highest occupied level of Carbon
Answer:
Highest occupied level of carbon is 3
Explanation:
its called Alkynes
Explanation:
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