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The answer is C. Hibernate during the cold winter months.
Why?
Alpine marmots are known for having a long hibernation duration which starts in October (winther) and ends in April (summer) (about 7 months). During this long period, they are able to reduce their bear beats from 200 per minute to just 30 or 38 beats, and their breaths from 60 breaths/minute to 1-3 breaths/minute, guaranteeing an extreme energy saving process.
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Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
Answer:
The coefficient before potassium (K) balances this chemical equation is 2.
Explanation:
_K +Cl₂ → 2KCl
K =1 ; Cl =2
K=1 × 2 = 2
Cl = 1 × 2 = 2
2 K +Cl₂ = 2 KCl
The pH of the solution after adding 0.150 moles of solid LiF is 3.84
<u>Explanation:</u>
We have the chemical equation,
HF (aq)+NaOH(aq)->NaF(aq)+H2O
To find how many moles have been used in this
c= n/V=> n= c.V
nHF=0.250 M⋅1.5 L=0.375 moles HF
Simillarly
nF=0.250 M⋅1.5 L=0.375 moles F
nHF=0.375 moles - 0.250 moles=0.125 moles
nF=0.375 moles+0.250 moles=0.625 moles
[HF]=0.125 moles/1.5 L=0.0834 M
[F−]=0.625 moles/1.5 L=0.4167 M
To determine the problem using the Henderson - Hasselbalch equation
pH=pKa+log ([conjugate base/[weak acid])
Find the value of Ka
pKa=−log(Ka)
pH=−log(Ka) +log([F−]/[HF]
pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)
pH=-log(3.5 x 10 ^4)+log(4.996)
pH= -4.54+0.698
pH=-(-3.84)
pH=3.84
The pH of the solution after adding 0.150 moles of solid LiF is 3.84