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uysha [10]
3 years ago
11

According to the following reaction, how many grams of water are produced in the complete reaction of 31.8 grams of barium hydro

xide?
Ba(OH)2(aq) + H2SO4(aq) --> BaSO4(s) + 2H2O(l)
Chemistry
1 answer:
Damm [24]3 years ago
7 0

Answer:

6.696 g of water (H₂O)

Explanation:

First we calculate the number of moles of barium hydroxide:

number of moles = mass / molecular weight

number of moles = 31.8 / 171.3 = 0.186 moles of barium hydroxide

Form the chemical reaction we may devise the following reasoning:

if         1 mole of barium hydroxide produces 2 moles of water

then   0.186 moles of barium hydroxide produces X moles of water

X = (0.186 × 2) / 1 = 0.372 moles of water

mass = number of moles × molecular weight

mass of water = 0.372 × 18 = 6.696 g

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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
4 years ago
EASYYYYY!!!!!!!!
nlexa [21]

crop rotation, green manure, and bone meal

Explanation:

I just looked it up. hope it helps

6 0
3 years ago
Convert 2.1 atm to kPa.
Ira Lisetskai [31]
212.782 kilopascals. Hope this helps :)
3 0
3 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
How many moles of H2 are in a flask with a volume of 2500 mL at a pressure of 30.0 kPa and a temperature of 27oC?
quester [9]

Answer:

0.0300 moles of H₂

Explanation:

The original equation is PV = nRT. We need to change this to show moles (n).

n = \frac{PV}{RT}

It's important to convert your values to match the constant (r) in terms of units.

30.0 kPa = 0.296 atm

2500 mL = 2.50 L

27 °C = 300 K

Now, plug those values in to solve:

n = \frac{(0.296)(2.50)}{(0.0821)(300)}    -  for the sake of keeping the problem clean, I didn't include the units but you should just to make sure everything cancels out :)

Finally, you are left with n = 0.0300 moles of H₂

4 0
3 years ago
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