When the guard cell is filled with water and it becomes turgid, the outer wall balloons outward, drawing the inner wall with it and causing the stomate to enlarge.
Answer:
1. Convert given mass of reactant to moles.
2. Convert moles of reactant to moles of product using the coefficients in the balanced equation.
3. Convert moles of product back to grams of product
Explanation:
In order to find out the mass of a given product in grams you would have to convert the grams into moles by multiplying 1 mole over the given mass of the reactant. Then, you would convert the moles of the reactant into moles of the product using mole ratio. And lastly, you would multiply by the mass of the product to calculate the mass back to grams.
Hope this helps!!<3
Answer:
The correct answer will be " RbF > RbCl > RbBr > Rbl".
Explanation:
The size of the given ions will be:
<u>RbCl:</u>
⇒ 689kJ/mol
<u>RbBr:</u>
⇒ 660kJ/mol
<u>Rbl:</u>
⇒ 630kJ/mol
<u>RbF:</u>
⇒ 785kJ/mol
Now according to the size, the arrangement will be:
⇒ (785kJ/mol) > (689kJ/mol) > (660kJ/mol) >(630kJ/mol)
⇒ RbF > RbCl > RbBr > Rbl
The bond among all opposite charging ions seems to be strongest whenever the ions were indeed small.
The mass of NaCl produced is 0.117 g.
<h3>What is the pH?</h3>
The pH of a solution is what tells us if the solution is acidic or basic. The solution is acidic if the pH of the solution is less than 7.
1) pH = - log(3.5 x 10^-5 M)
= 4.45
Thus the solution is acidic.
2) We can see that the concentration as given is the concentration of OH^-
[OH ] = 4.0 x 10^-3 M
pOH = -log (4.0 x 10^-3 M)
pOH = 2.39
pH = 14 - 2.39
pH = 11.61
The equation of the reaction is; 
Number of moles of NaOH = 0.40 M * 10/1000 = 0.004 moles
Number of moles of HCl = 0.20 M * 10/1000 = 0.002 moles
We can see that HCl is the limiting reactant
Mass of NaCl produced ;
0.002 moles * 58.5 g/mol = 0.117 g
Learn more about reaction:brainly.com/question/3588292
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Answer is: mass of methanol is 528.32 grams.
N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.
n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).
n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.
n(CH₃OH) = 16.49 mol; amount of substance.
m(CH₃OH) = n(CH₃OH) · M(CH₃OH).
m(CH₃OH) = 16.49 mol · 32.04 g/mol.
m(CH₃OH) = 528.32 g.
