2. How many orbitals are in the following sublevels?

2. What ions are present in what ratio in a solution of aqueous calcium chloride?

Alenkasestr [34]

Answer:

$\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }$

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form $\mathrm{Ca}^{2+}$ , while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two $\mathrm{Cl}^{-}$ ions.

$\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}$

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one $\mathrm{Ca}^{2+}$ and two $\mathrm{Cl}^{-}$ ions: $\mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})$

5 0

3 years ago

What volume of oxygen at STP is requieres for the complete combustion of 100.50 mL of C2H2

malfutka [58]

The volume of oxygen at STP required would be 252.0 mL.

<h3>Stoichiometic problem</h3>

The equation for the complete combustion of C2H2 is as below:

$2C_2H_2 + 5O_2 --- > 4CO_2 + 2H_2O$

The mole ratio of C2H2 to O2 is 2:5.

1 mole of a gas at STP is 22.4 L.

At STP, 100.50 mL of C2H2 will be:

100.50 x 1/22400 = 0.0045 mole

Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles

0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL

Thus, 252.0 mL of O2 gas will be required at STP.

More on stoichiometric problems can be found here: brainly.com/question/14465605

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7 0

1 year ago

H-E-L-P!!...!! ʕ ꈍᴥꈍʔ

N76 [4]

Answer:

the correct answer is (2)

Explanation:

4 0

2 years ago

Read 2 more answers

In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate th

Leto [7]

Answer:

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu(NO3)2*3H2O

Explanation:

Step 1: Data given

Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

Step 3: Calculate moles of Cu(NO3)2

Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2

Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol

Moles Cu(NO3)2 = 0.144 moles

Step 4: Calculate moles of H2O

Moles H2O = 7.7892 grams / 18.02 g/mol

Moles H2O = 0.432 moles

Step 5: Calculate Z

z = moles H2O / moles Cu(NO3)2

Z = 0.432/0.144

Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O

5 0

3 years ago

The temperature at which water vapor condenses into liquid water

Marysya12 [62]

Precipitation

100°C
When 1 mol of water vapor at 100°C condenses to liquid water at 100°C, 40.7 kJ of heat are released into the surroundings.

3 0

2 years ago