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bogdanovich [222]
3 years ago
10

How many atoms are there in 5.70 mol of hanfnium?

Chemistry
1 answer:
ANTONII [103]3 years ago
6 0

The correct answer would be 3.49 times 10^ minus 24 molecules

You might be interested in
The central Xe atom in the XeF4 molecule has ________ unbonded electron pair(s) and ________ bonded electron pair(s) in its vale
Dimas [21]

Answer:

Lewis structure in attachment.

Explanation:

Atoms of elements in and beyond the third period of the  periodic table form some compounds in which more than eight electrons surround the  central atom. In addition to the 3s and 3p orbitals, elements in the third period also  have 3d orbitals that can be used in bonding. These orbitals enable an atom to form  an <u>expanded octet</u>.

The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.

8 0
3 years ago
A polar covalent bond will form between which two atoms?
Alex_Xolod [135]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)


For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

                                                   ________

                    E.N Difference                0.96          (Polar Covalent Bond)


For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

                    E.N Difference                2.51          (Ionic Bond)


For F and F,

                    E.N of Fluorine          =   3.98

                    E.N of Fluorine          =   3.98

                                                        ________

                    E.N Difference                0.00         (Non-Polar Covalent Bond)

Result:

           A polar covalent bond is formed between Hydrogen and Chlorine atoms.

5 0
3 years ago
Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles o
Harman [31]
First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹

The answer is C
8 0
3 years ago
Tin (II) fluoride, formerly found in many kinds of toothpaste, is formed in this reaction: Sn (s) + 2HF (g) ——&gt; SnF2 (s) + H2
Readme [11.4K]

1.34 L of HF

Explanation:

We have the following chemical reaction:

Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)

First we calculate the number of moles of SnF₂:

number of moles = mass / molecular weight

number of moles of SnF₂ = 5 / 157 = 0.03 moles

From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.

Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:

number of moles = volume / 22.4 (L/mole)

volume of HF = number of moles × 22.4

volume of HF = 0.06 × 22.4 = 1.34 L

Learn more about:

problems with gases at STP

brainly.com/question/8857334

#learnwithBrainly

8 0
3 years ago
A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th
Kisachek [45]

Answer:

pH=4.1

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1

Regards.

3 0
3 years ago
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