What is the interval used on the y-axis of this graph?

What is the definition of a chemical bond?

sweet-ann [11.9K]

Answer:

B- A mutual attraction between the nuclei and the electrons in two different atoms

4 0

2 years ago

Calculate the number of atoms in 3 moles of carbon atoms

creativ13 [48]

Answer:

1,806,000,000,000,000,000,000,000 atoms.

Explanation:

1 mol of anything is 6.022*10^23 units

1 mol carbon atoms = 6.022*10^23 carbon atoms

3 mol carbon atoms = 3*6.022*10^23 = 1.8*10^24 atoms

Correctly - your answer should have only 1 significant digit: 2 * 10^24 atoms.

6 0

1 year ago

How many atoms are in 0.100 mol of C3H8

Natalija [7]

Number of atoms = mole x Avogadro’s Number

= 0.1 x 6.022 x 10^23
= 6.022 x 10^22

8 0

2 years ago

When 3243. grams of iron (III) chloride are reacted with 511.8 grams of hydrosulfuric acid, which is the limiting reactant?

AleksAgata [21]

Answer:

Hydrosulfuric acid will act as limiting reactant.

Explanation:

Given data:

Mass of iron(III) chloride = 3243.0 g

Mass of hydrosulfuric acid = 511.8 g

Limiting reactant = ?

Solution:

Chemical equation:

2FeCl₃ + 3H₂S → Fe₂S₃ + 6HCl

Number of moles of iron(III) chloride:

Number of moles = mass/molar mass

Number of moles = 3243.0 g/ 162.2 g/mol

Number of moles = 20 mol

Number of moles of hydrosulfuric acid:

Number of moles = mass/molar mass

Number of moles = 511.8 g/ 34.1 g/mol

Number of moles = 15 mol

Now we will compare the moles of both reactant with products

FeCl₃ : Fe₂S₃

2 : 1

20 : 1/2 ×20 = 10

FeCl₃ : HCl

2 : 6

20 : 6/2 ×20 = 60

H₂S : Fe₂S₃

3 : 1

15 : 1/3 ×15 = 5

H₂S : HCl

3 : 6

15 : 6/3 ×15 = 30

Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.

5 0

2 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0

2 years ago

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