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Romashka-Z-Leto [24]
3 years ago
11

How does pressure affect an organism plants?

Chemistry
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

Turgor pressure in plants. Turgor pressure within cells is regulated by osmosis and this also causes the cell wall to expand during growth.

Explanation:

I hope that helps

You might be interested in
Attempt 3 During an experiment, a student adds 2.90 g CaO 2.90 g CaO to 400.0 mL 400.0 mL of 1.500 M HCl 1.500 M HCl . The stude
kondor19780726 [428]

Answer:

Explanation:

Equation of the reaction:

CaO(s) + 2H+(aq) -----> Ca2+(aq) + H2O(g)

The ∆Hrxn would be for one mole of CaO reacted or 2 moles of H+, whichever is the limiting reactant.

Number of moles = mass ÷ molar mass

Molar mass of CaO = 40 + 16

= 56 g/mol

moles of CaO = 2.90/56

= 0.0518 mol

Number of moles = concentration × volume

moles of HCl = 400 × 10^-3 × 1.500 = 0.6 moles

Moles of HCl = moles of H+

From the equation, 1 mole of CaO reacted with 2 moles of H+ to give 1 mole of water.

To find the limiting reagent,

0.6 mole of H+/2 moles of H+ × 1 mole of CaO

= 0.3 moles of CaO(> 0.0518 moles)

So, CaO is limiting reactant.

∆H = m × Cp × ∆T

m = density × volume

= 400 × 1

= 400 g

Cp = 4.184 J/g-ºC

∆T = +6 ºC

∆H = 400 × 4.184 × 6

= 10041.6 J

Since the reaction is exothermic,

∆Hrxn = -∆H/mol(CaO)

= -10041.6/0.0518

= -193853 J

= -193.9 kJ/mol.

3 0
3 years ago
How do I do this? What do I do what are the steps I have no idea
Vesna [10]
B should be 20 because only 1 oxygen goes into water
C is hydrogen because 2 hydrogens go into each water
Can't read anything beyond c
7 0
3 years ago
Read 2 more answers
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

3 0
4 years ago
Scientist A produces 83.67 g KMnO4 while Scientist B produces 81.35 g KMnO4.
noname [10]

Answer:

Y_A=92.1\%\\\\Y_B=89.6\%

Explanation:

Hello there!

In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:

m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2}  *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4

Now, we are able to compute the percent yields, by using the actual yield each scientist got:

Y_A=\frac{83.67g}{90.9g} *100\%=92.1\%\\\\Y_B=\frac{81.35g}{90.9g} *100\%=89.6\%

Regards!

8 0
3 years ago
4. What is the percent by mass of tungsten in the calcium tungstate compound CaWO3, commonly
MA_775_DIABLO [31]

Answer: 67.4%

Explanation:

Ca 40.8

W. 183.8

O. 16.00

CaWO3 272.6

W = 183.8/272.6 = 0.674

3 0
3 years ago
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