Answer:
[CaCl₂] = 1.32 M
Explanation:
We know the volume of solution → 0.30 L
We know the mass of solute → 44 g of CaCl₂
Let's convert the mass of solute to moles.
44 g . 1 mol / 110.98 g = 0.396 moles
Molarity (mol/L) → 0.396 mol / 0.3 L = 1.32 M
Answer:
it will probably flame up or explode or maybe start boiling
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>
First, the equation of the reaction:

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
T2 = 51.6°C
Explanation:
Given:
P1 = 1.01 atm
T1 = 25°C + 273 = 298K
P2 = 1.10 atm
T2 = ?
P1/T1 = P2/T2
Solving for T2,
T2 = (P2/P1)T1
= (1.10 atm/1.01 atm)(298K)
= 324.6 K
= 51.6°C
where Tc = Tk - 273
The concentration of the HCl solution is 0.047 M.
Explanation:
Data given about acid and base:
volume of acid Vacid = 46.9 ml
molarity of acid =?
volume of the base (NaOH) = 16.4 ml
molarity of the base = 0.135 M
To know the concentration of the acid in this reaction, the formula used in titration is used. It is
Macid X Vacid = Mbase X Vbase
the formula is rewritten as:
Macid = 
putting the values in the equation:
Macid = 
= 0.047 M
the concentration of the acid i.e HCl in the solution is of 0.047 M.