The dissolution<span> of </span>calcium chloride<span> in </span>water<span>. </span>CaCl2(s<span>)--> </span>Ca^2+ (aq) +2 Cl^-1<span> (</span>aq<span>). A) </span>Use<span> the </span>data<span> from</span>appendix C<span>. and </span>below<span> to </span>determine Delta H<span> of </span>formation<span> in </span>kj/mol<span> for this </span>process<span> (</span>deltaH(Ca^2+(aq)=-542.8 kj/mol<span>).</span>
Answer:
a) <u>1.740 g</u> of F- must be added to a cylindrical water reservoir
b) Grams of sodium fluoride, NaF, that contain this much fluoride:
3.84 g
Explanation:
Step 1. calculate the volume of the tank:
Volume of cylinder =
,
Here r = radius of the cylinder = d/2
h = depth = 21.80m


= 168 m
Volume =


2.Convert ppm to g/m3 and Solve for mass of F-


Because both ppm and g/m3 are same quantity .



mass of F- required = 1.740 g
3. Apply <u>mole concept </u>to calculate grams of sodium fluoride produced
mass of 1 mole of F2 = 38 g
mass of 1 mole of NaF = 42 g
(from periodic table calculate molar mass)

Here 1 mole of F2 produce = 2 mole of NaF
So,
38 g of F2 produce = 2 x 42 g of NaF
38 g of F2 produce = 84 g of NaF
1 g of F2 produce = 84/38 g of NaF
1.74 g F2 produce =

1.74 g F2 produce = 3.84 g of NaF
3.84 g of NaF is produced
Answer:
A
Explanation:
To calculate questions like this, you need to make use of the equation below:
E = hν
E is the energy, h is Planck's constant, and ν is the frequency.
Also, another important formula you need for this question is the following:
ν = c / λ
ν is the frequency, c is the speed of light, and λ is the wavelength.
Therefore, E = hν = hc/λ.
Also remember to convert nm wavelength to meters.
To answer the question, E = hc/λ = ((6.626*10^-34 J•s)(3*10^8 m/s))/(4.45*10-7 m) = 4.42 • 10^-19 J = A
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
By there pH . a pH below 7 is acidic . Above 7 is basic. If it’s right at 7 it’s neutral.