Answer:
1 no. herbivore ....... 2 no. exception......3 no. diameter........4 no. heart........ 5 no. catastrophe...... 6 no. develop...... mark me brainliest plzzzz
A decrease in population which means animals and plants.
Answer:
The precipitate will form.
Explanation:
Let's write the equilibrium expression for the solubility product of calcium sulfate:
⇄ 
The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:
![K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D10%5E%7B-4.5%7D)
Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:
![Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}](https://tex.z-dn.net/?f=Q_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D5.00%5Ccdot10%5E%7B-2%7D%20M%5Ccdot7.00%5Ccdot10%5E%7B-3%7D%20M%3D3.5%5Ccdot10%5E%7B-4%7D)
Now given the K value of:

Notice that:

This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.
Answer:
401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.
Explanation:
The expression for the standard change in free energy is:
Where,
is the change in the Gibbs free energy.
T is the absolute temperature. (T in kelvins)
is the enthalpy change of the reaction.
is the change in entropy.
Given at:-
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
= 128.9 kJ/mol
= 33.1 kJ/mol
Applying in the above equation, we get as:-

= 0.32131 kJ/Kmol
So, For reaction to be spontaneous, 
Thus, For minimum temperature:-

<u>Hence, 401.17 K is the minimum temperature at which the reaction will become spontaneous under standard state conditions.</u>
Answer:
Anhydrous calcium chloride dissolves and becomes liquid
Anhydrous copper (ii) sulphate will produce crystal particles
Explanation:
Anhydrous calcium chloride is deliquescent and hence when it is exposed to air, it absorbs water from air. After absorbing water, it dissolves and after some time a pool of clear liquid appears.
Anhydrous copper (ii) sulphate will form crystal structures and the following reaction will takes place
CuSO4 + 5 H20 --> CuSO4.5H2O