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Andre45 [30]
3 years ago
6

At 1 atm of pressure and a temperature of 0 degrees celsius, which phases of H2O can exist?

Chemistry
1 answer:
NemiM [27]3 years ago
3 0
B, because water is changing its state from solid to liquid (it's fusion in portuguese, don't know in english), so while it's changing, water has 2 states at the same time.
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if the pressure of a gas at constant volume is 3.5 atm at 100°c, what will the pressure be if the tempature is changed to 250°c?
Degger [83]
Sorry don't know this one

6 0
4 years ago
Neon doesn't take part in chemical reaction why give reason​
ankoles [38]
Neon has filled its outer shells, therefore it is very stable and does not need to react with other elements and doesn’t form compounds.
5 0
3 years ago
Read 2 more answers
Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.
melomori [17]

Answer:

0.15M

Explanation:

The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.

You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, \frac{25.0g}{1} × \frac{1 mol}{56.106g} = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

The volume is given in L already so no need to do any conversion. M= \frac{0.4558mol}{3.00L} = 0.1485M ≈ 0.15M

5 0
2 years ago
6. According to Crystal Field Theory, what should be the color of an aqueous solution of the ion [Y(H2O)6]3
UkoKoshka [18]

Answer:

Colourless

Explanation:

We know that Y^3+ has the electronic configuration of;

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (the 5s and 4d levels are empty).

According to the crystal field theory, the colour of complexes result from transitions between incompletely filled d orbitals.

As a result of this, complexes with empty or completely filled d orbitals are colourless. Thus,  [Y(H2O)6]3 is colourless according to the Crystal Field Theory.

4 0
3 years ago
A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
OLEGan [10]

Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

6 0
3 years ago
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