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Ilya [14]
3 years ago
7

10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?

Chemistry
1 answer:
il63 [147K]3 years ago
8 0
Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
  ? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!
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Calculate the pH of a solution that is 0.205M benzoic acid and 0.230M sodium benzoate, a salt whose anion is the conjugate base
pickupchik [31]

Answer:

pH = 4.25

Explanation:

A solution composed of a weak acid and its conjugate base is a <em>buffer solution</em>. To calculate the pH of a buffer solution we use the Henderson-Hasselbach equation:

pH = pKa + log ([conjugate base]/[weak acid]

In this case, we have the following data:

[conjugate base] = [sodium benzoate] = 0.230 M

[weak acid] = [benzoic acid] = 0.205 M

The pKa of benzoic acid is 4.2. So, we introduce the data in the equation:

pH = 4.2 + log (0.230 M/0.205 M) = 4.2 + 0.050 = 4.25

4 0
2 years ago
Which best describes the products of a chemical reaction?
cricket20 [7]

Answer:

Q1 part D Q2 part B

Explanation:

After the chemical reaction, the things that you get are known as products.

3 0
2 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
Approximately how many atoms make up 2.3 moles of gallium?
Len [333]
I think its b I think
5 0
3 years ago
Write the appropriate symbol for each of the following isotopes: (a) Z 74, A 186; (b) Z 80, A 201; (c) Z 34, A 76; (d) Z 94, A 2
beks73 [17]

Answer:  a)  _{74}^{186}\textrm{W}

b) _{80}^{201}\textrm{Hg}

c)  _{34}^{76}\textrm{Se}

d) _{94}^{239}\textrm{Pu}

Explanation:

General representation of an element is given as:_Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

In an atom, when neutrons or protons are lost or gains, it directly affects the mass number of an atom.

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

It is characteristic of a particular element.

Atomic number = Number of electrons = Number of proton

a) Z 74, A 186: _{74}^{186}\textrm{W}

b) Z 80, A 201: _{80}^{201}\textrm{Hg}

c) Z 34, A 76: _{34}^{76}\textrm{Se}

d) Z 94, A 239.: _{94}^{239}\textrm{Pu}

6 0
3 years ago
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