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4 years ago

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

1 answer:

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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When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the

klemol [59]

Answer:

The equivalent circuit for the electrode while the electrolyte gel is fresh

From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin

Now as the electrolyte gel start to dry out the resistance $R_s$ of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out the resistance of the gel $R_s$ then increases to infinity and this in turn cut off flow of current.

The diagram illustrating this is shown on the second uploaded image

Explanation:

5 0

3 years ago

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

tensa zangetsu [6.8K]

<h3>Answers:</h3>

1) 2 Units of Ozone

2) 3 Units of Ozone

3) 9 Units of Ozone

<h3>Solution:</h3>

1) From 6 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

6 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

9 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 3 Units of Ozone

3) From 27 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

27 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 9 Units of Ozone

3 0

3 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

<u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0

3 years ago

WILL GIVE BRAINLIST

8_murik_8 [283]

I would choose C bc if u look at it closely u can notice it goes up steadily and then BANG it decreases a lot

4 0

3 years ago

How many atoms of chlorine (Cl) are shown below? <br><br> 5 CaCl2

Jobisdone [24]

The answer is not C...

5 0

4 years ago