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statuscvo [17]
2 years ago
13

Ice is actually frozen ______

Chemistry
2 answers:
sergeinik [125]2 years ago
8 0

Answer:

water

Explanation:

or h2o i mean theyre the same thing soo..

Ann [662]2 years ago
3 0
H20 (water), hope this helps
You might be interested in
Now, Johnny the pool guy must add enough Na2CO3 to get the pH back up to 7.60. How many grams of Na2CO3 does Johnny need to add
vaieri [72.5K]

Answer:

m = 3.4126 g

Explanation:

First, the question is incomplete but I already put in the comments the rest of the question.

Let's solve the first two questions, and then the actual question you are asking here to give you a better explanation of how to do it.

1) We need the volume of the pool, in this case is easy. Assuming the pool is rectangular, we use the volume of a parallelepiped which is the following:

V = h * d * w

We have the data, but first we will convert the feet to centimeter. This is because is easier to work the volume in cm³ than in feet.

So the height, width and depth of the pool in centimeter are:

h = 32 * 30.48 = 975.36 cm

w = 18 * 30.48 = 548.64 cm

d = 5.3 * 30.48 = 161.54 cm

Now the volume:

V = 975.36 * 548.64 * 161.54

V = 86,443,528.79 cm³ or 86,443,528.79 mL or 86,443.5 L

2) If the pool has a pH of 6.4, the concentration of H+ can be calculated with the following expression:

[H+] = antlog(-pH) or 10^(-pH)

Replacing we have:

[H+] = 10^(-6.4)

[H+] = 3.98x10^-7 M

3) Finally the question you are asking for.

According to the reaction:

Na2CO3 + H+ → 2Na+ + HCO3−

We can see that there is ratio of 1:1 between the H+ and the Na2CO3, so, if we have initially a concentration of 3.98x10^-7 M, the difference between the new concentration of H+ and the innitial, will give the concentration to be added to the pool to raise the pH. Then, with the molecular weight of Na2CO3 (105.98 g/mol) we can know the mass needed.

The new concentration of [H+] is:

[H+] = 10^(-7.6) = 2.58x10^-8 M

The difference of both [H+] will give concentration of Na2CO3 used:

3.98x10^-7 - 2.58x10^-8 = 3.73x10^-7 M

The moles:

moles = 3.73x10^-7 * 86,443.5 = 0.0322 moles

Finally the mass:

m = 0.0322 * 105.98

m = 3.4126 g    

5 0
3 years ago
How many moles of Oxygen are in 0.52 mol of Ca(NO3)2?
RideAnS [48]
2 moles of NO3 contains 6 moles of O
8 0
3 years ago
Where does the most geologic activity ( including volcanos and earthquakes ) occur
Gekata [30.6K]
I believe it occurs along the plates edge.
3 0
3 years ago
Read 2 more answers
How many moles of nitrogen, N, are in 0.286 mil of ammonium nitrate?
Arisa [49]

Answer:

0.572

Explanation:

The formula for ammonium nitrate is

NH4NO3

The trick here is to realize the 4 applies only to the hydrogen. The 3 applies only to the oxygen.

There are 2 nitrogens in the formula. So if the entire molecule represents 0.286 mols. The individual mols of each part of the formula is just multiplied by the number of moles of the molecule. In formula form

Number of moles of the atoms making up the molecule = number of atoms from the formula * mols of the molecule.

2*0.286 = 0.572 moles of Nitrogen

8 0
3 years ago
In another experiment, the student titrated 50.0mL of 0.100MHC2H3O2 with 0.100MNaOH(aq) . Calculate the pH of the solution at th
Helen [10]

pH=8.87

<h3>Further explanation</h3>

Reaction

C₂H₄O₂+NaOH⇒CH₃COONa+H₂O

at the equivalence point = mol C₂H₄O₂= mol NaOH

mol C₂H₄O₂ : 50 x 0.1 = 0.5 mlmol=5.10⁻⁴ mol

The two reactants have completely reacted, and there is only salt(CH₃COONa) and water(H₂O), there will be hydrolysis

For acids from weak acids and strong bases (the solution is alkaline) then the calculation:

\tt [OH^-]=\sqrt{\dfrac{Kw}{Ka}\times M }

M=anion concentration=CH₃COO⁻

Ka=acid constant(for CH₃COOH,Ka=1.8.10⁻⁵)

\tt [OH^-]=\sqrt{\dfrac{10^{-14}}{1.8.10^{-5}}\times 0.1 }

\tt [OH^-]=\sqrt{5.6.10^{-11}}=7.483\times 10^{-6}

pOH=6-log 7.483=5.13

pH= 14 - 5.13=8.87

5 0
3 years ago
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