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Answer:
Explanation:
what are tho anwser choices
Related Questions
Explain how atoms organize to create larger structures?
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Answer:
First of all they are super tiny particles of neutrons, electrons and protons. Then they come together to organize a structure of atom. Atoms come together to form a molecule and smaller molecules work together to form macromolecules. Thus, how they organize and form building blocks of a substance.
Hope it helps!<3
How many joules are required to melt 250 grams of water
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Answer:
i not sure but i searched it up and it said Q=104525J
Explanation:
Pro
∆G0=-RT in k Known as
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Answer:
Gibbs free energy equation
Explanation:
A metal ion uses d2sp3 orbitals when forming a complex. What is its coordination number and the shape of the complex?
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Answer:
Octahedral
Explanation:
The idea of hybridization is a valence bond concept introduced by Linus Pauling and soon became an important explanation for the structure of molecules even in coordination chemistry.
Hybridization involves the mixing of orbitals of appropriate energy to form a new set of orbitals that are more suitable for bonding with ligands.
When one s, two d and three p orbitals are hybridized in complex formation, octahedral complexes are formed.
A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution?
A) 0.126g
B) 3.66g
C) 4.03g
D) 3.84g
E) 280 g
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Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
An ice cube melts on the warm sidewalk on a hot summer day. Let the entropies of the ice cube, of the pavement and of the ice cubesidewalk system be Sice, Ssw and Ssys. What happens to these entropies?
1. Sice increases, Ssw decreases, Ssys does not change.
2. Sice decreases, Ssw increases, Ssys decreases
3. Sice increases, Ssw decreases, Ssys increases
4. Sice increases, Ssw increases, Ssys increases
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Answer:
3. Sice increases, Ssw decreases, Ssys increases
Explanation:
Entropy basically refers to the degree of disorderliness of a system.
Ice Cube --> As the ice cube melts, it undergoes a change in state of matter from solid to liquid hence it's entropy increases as the degree of movement of its molecules is more disordered in the liquid state.
This means the likely options are; 1, 3 and 4.
Sidewalk --> The entropy here decreases as the ice melts on the side walk. This is because there is a reduction in temperature. Entropy is directly proportional to temperature. Decrease in temperature leads to decrease in entropy.
This means the likely option is 1 and 3
The net entropy change of the system is however positive.
Polonium-214 went through a series of radioactive decays to produce a stable isotope of bismuth-210 choose the correct decay series that would produce bismuth-210
A: Polonium-214 undergoes 2 beta decay and 1 gamma emission
B: Polonium-214 undergoes 1 alpha decay and 2 gamma emission
C: Polonium-214 undergoes 2 alpha decay and 1 beta decay
D: Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.
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The correct decay series :
₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰
Further explanationRadioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,
alpha α particles ₂He⁴ beta β ₋₁e⁰ particles gamma particles ₀γ⁰ positron particles ₁e⁰ neutron ₀n¹Decay's reaction :
₈₄Po²¹⁴⇒₈₃Bi²¹⁰+x+y+z
From this decay reaction can be concluded
mass number decreased by 4
atomic number decreased by 1
Then the most suitable are:
Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.
₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰
So that the mass number and atomic number are the same both before and after decay
