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Katarina [22]
2 years ago
11

K3PO4(aq)+MgCl2(aq)⟶

Chemistry
1 answer:
Wittaler [7]2 years ago
7 0

Answer:

6KCl(aq) +  Mg3(PO4)2(s)  ...  balanced equation

Explanation:

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A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand throug
djyliett [7]

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>

<em></em>

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>

<em></em>

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial  = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

7 0
3 years ago
Balance the equation for ethane C3H6 burning in oxygen to form carbon dioxide and steam. _____C3H6 + ____ O2 ---&gt; CO2 + ____H
aleksandrvk [35]

Answer:

   2C3H6 + 9 O2 ---> 6 CO2 + 6 H2O

Explanation:

8 0
3 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
How to keep pipes from freezing during power outage?.
Maurinko [17]

Answer: <em>Running water, even at a trickle, helps prevent pipes from freezing. If you decide to use fuel-burning equipment, such as a kerosene heater, ensure it is vented to outside and kept clear of any vents. Keep garage doors closed if there are water supply lines in the garage.</em>

Explanation:

8 0
2 years ago
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How is hydrogen different from alkali metals?
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Hydrogen is different from alkali metals because it also exhibits the properties of inert gas
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