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3 years ago

11

A herd of sheep moves up the mountain to graze during summer, and migrates to the valley during winter. What type of migration d

o the sheep follow

2 answers:

frosja888 [35]3 years ago

6 0

<span>its seasonal movements</span>

agasfer [191]3 years ago

5 0

<h2>Answer:</h2>

This type of migration is known as the seasonal migration.

<h3>Explanation:</h3>

Seasonal migration means moving from one place to another due to the change in seasons.
Cattles like sheeps move to mountains in summer to graze because in summer the temperature of the mountains is less. So sheeps move to mountains to escape heat.
In winters, sheeps move to valley. Because in winters, there are no fields or plants on mountains due to snow.

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Ignore the answer I have selected but please help!! Ty in advance, also please explain

weeeeeb [17]

Answer:

Iodine is most reactive because it is very close to having a "full shell" which is 8 electrons so they are "eager" to gain the last electron to became balanced, so that makes it the most reactive. Hope that helps:)

Explanation:

4 0

2 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r

son4ous [18]

Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

5 0

3 years ago

Lab 27. stoichiometry and chemical reactions: which balanced chemical equation best represents the thermal decomposition of sodi

Cloud [144]

Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.

<span> 2 NaHCO</span>₂<span> </span> →<span> Na</span>₂<span>CO</span>₃<span> (s) </span>+ <span> CO</span>₂<span> (g) + H</span>₂<span>O (g)
</span>
Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.

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2 years ago

Give a chemical equation for aerobic cellular respiration

Aleonysh [2.5K]

C6H12O + 6OC2 + 6H2O + energy

5 0

3 years ago