All categories

2 years ago

Does the amount of water affect the growth of a plant?

Chemistry

2 answers:

GREYUIT [131]2 years ago

3 0

Answer:

Yes

Explanation:

It does because of the fact that plants need water to live and the grow. Water helps the plant grow, and certain plants would need less or more water based on their roots, leaves, and tempurature tolerancy.

babunello [35]2 years ago

3 0

Answer: Yes,

Roots need both water and oxygen, and when surrounded by water, they cannot take up oxygen. Plants may slow in growth after a flush of new growth or a heavy flowering. During these periods and while it is dormant, a plant will need less water.

You might be interested in

Acidic soil with a surface of decayed pine needles describes the

Sedbober [7]

Allelopathy, the pine needles make the ground more acidic so weeds cannot grow and steal nutrients the tree needs

7 0

3 years ago

The colour imparted to a flame by calcium ion?

igor_vitrenko [27]

I not sure, but i know that is soft yellow no white not yellow.

7 0

3 years ago

The process of converting a liquid to a gas is called

Gekata [30.6K]

Converting a liquid to a gas is evaporation.

6 0

3 years ago

Read 2 more answers

A 13.9 - g piece of metal ( specific heat capacity is 0.449 /g^ C)who whose temperature is 54.2 degrees * C was added to a sampl

lubasha [3.4K]

Answer:

26.2g = Mass of water in the calorimeter

Explanation:

The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:

Q = m*C*ΔT

Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature

Replacing we can write:

$m_{metal}*C_{metal}*dT_{metal}=m_{water}*C_{water}*dT_{water}$

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)

240.9J = m(H₂O) * 9.2J/g

<h3>26.2g = Mass of water in the calorimeter</h3>

4 0

3 years ago

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago