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vaieri [72.5K]
3 years ago
14

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00- mL sam

ple of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium.
Chemistry
1 answer:
noname [10]3 years ago
8 0

Answer:

Concentration ammonium is 0.036 g/mL

Explanation:

Stock solution has concentration of 0.108g/mL. Take 10mL stock solution and add 50 mL of water meaning it has 1,08g/60mL. Ammonium sulfate has 2 molecules of cation ammonium cation. So, (1,08g/60mL.)*2=0.036 g/mL

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jolli1 [7]

Answer:

Number of protons and neutrons

7 0
3 years ago
3 and
kenny6666 [7]

Answer:

Hey I dont know it but good luck!! :)

Explanation:

3 0
3 years ago
Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
Equal volumes of hydrogen and helium gas are at the same pressure. The atomic mass of helium is four times that of hydrogen. If
OverLord2011 [107]

Answer:

The ratio of the temperature of helium to that of hydrogen gas is 2:1.

Explanation:

Atomic mass of hydrogen = M

Temperature of hydrogen gas =T

Pressure of the hydrogen gas = P

Mass of the hydrogen gas = m

Moles of the hydrogen gas = n=\frac{m}{2M}

Volume of the hydrogen gas = V

Using an ideal gas equation:

PV=nRT=PV=\frac{mRT}{M}...(1)

Temperature of helium gas =T'

Pressure of the helium gas = P'= P

Mass of the helium gas = m' =m

Moles of the helium  gas = n'=\frac{m}{M'}=\frac{m}{4M}

Volume of the helium gas = V' = V

Using an ideal gas equation:

P'V'=n'RT'=\frac{mRT'}{4M}...(2)

Divide (2) by (1)

\frac{P'V'}{PV}=\frac{\frac{mRT'}{4M}}{\frac{mRT}{2M}}

\frac{T'}{T}=\frac{2}{1}

The ratio of the temperature of helium to that of hydrogen gas is 2:1.

7 0
3 years ago
Upon arrival we needed to hunt in this new land we only had five refills and they needed 50 g of gunpowder to be shot once. We o
emmasim [6.3K]

Answer:

Explanation:

Upon arrival we needed to hunt in this new land we only had five refills and they needed 50 g of gunpowder to be shot once. We only have 15 pounds of gunpowder. It is taking six shots to kill one of these wild turkeys. How many turkeys can be shot with 15 pounds of gunpowder?

If we had plenty of refills, and it takes 6 shots to kill a wild turkey at 50 gms of gunpowder per shot, then each turkey requires 6X50 =300gms of gunpowder.  We have 15X454 gms of gunpowder and have the potential to kill 15X454/300=22.7 or 22 turkeys.and it takes 6 shots to kill a wild turkey.

The limiting reagent is the number of refills, and withonly 5, we are out of luck and can't kill one turkey

6 0
2 years ago
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