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tigry1 [53]
3 years ago
12

For SCl4, the electron domain geometry is _______(i)________ and the molecular geometry is ______(ii)________.

Chemistry
1 answer:
Likurg_2 [28]3 years ago
8 0

The electron domain geometry is trigonal bipyramidal while the molecular geometry of the compound is seesaw.

The shapes of molecules is determined by the number of electron pairs on the valence shell of the central atom in the molecule. These electron domains include lone pairs and bond pairs.

The lone pairs only contribute towards the electron domain geometry and not the molecular geometry. SCl4 has five electron domains hence its electron domain geometry is trigonal bipyramidal. The molecular geometry of the compound is seesaw.

Learn more: brainly.com/question/6505878

You might be interested in
Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat cont
dsp73

Answer:

The products have a higher heat content than the reactants.

Explanation:

The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.

5 0
3 years ago
Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation AZX. Express your answer as a
Lelechka [254]

Answer:

^{32}_{15}P

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

However, when we talk about the atomic number of the ion, it is not equal to the number of electrons as electron can be gained or loosed.

This is why, more appropriately, the number of the protons which are present in the nucleus of the atom is called the atomic number.

Thus, atomic number of phosphorus = 15

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Given, Mass number = 32

Thus, the symbol of the isotope is:-

^{32}_{15}P

8 0
3 years ago
How many atoms of hydrogens are found in 3.21 mol of<br>C3H8?​
damaskus [11]

Answer:

1.55 × 10²⁵ atoms of H  

Explanation:

3.21mol C₃H₈ × 8mol H × (6.022×10²³)

8 0
3 years ago
How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?
zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\  =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\

2) Determine the limitting reagent

\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

6 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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