During a storm, a tree fell over into a river. What might happen to this tree?

How many atoms of phosphorus are in 8.10 mol of copper(II) phosphate?

madreJ [45]

You know from the chemical formula that 1 mole of Cu3(PO4)2 contains 2 moles of P so you can work out how many moles of P are in 8.1 moles of Cu3(PO4)2.

As for anything, 1 mole of X contains 6.022 * 10^23 of X. Multiply moles of X by Avogadro's number.

7 0

2 years ago

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Best answer and explanation will get marked brainiest

Darina [25.2K]

The number 345,000 has 3 significant figures because the trailing zeros are not significant since there's no decimal point.

<h3>What is significant figures?</h3>

Significant figures of a number in positional notation are digits in the number that are reliable and necessary to indicate the quantity of something.

Also significant figures can be defined as, the number of digits in a value, often a measurement, that contribute to the degree of accuracy of the value.

Examples of significant figures;

405 = 3 significant figures
405000 = 3 significant
0.040500 = 5 significant figures

Note: leading zeros are not significant but trailing zeros, which are zeros at the end of a number, are significant only if the number has a decimal point.

345,000 = 3.45 x 10⁵ (3 significant figures)

Thus, the number 345,000 has 3 significant figures because the trailing zeros are not significant since there's no decimal point.

Learn more about significant figures here: brainly.com/question/24491627

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4 0

2 years ago

10. What is the name of the acid with the formula HCIO?

pickupchik [31]

Answer:

THE NAME OF THE ACID WITH FORMULA HCIO IS HYPOCHLOROUS ACID

3 0

2 years ago

I need help, someone please help me

MArishka [77]

Answer:

c) the panda eats both plant and animal: fish and bamboo

d)it shows direction of feeding, the one below is fed on by the one above

6 0

2 years ago

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Can someone help on five and six I am at a loss on how to solve it

Bumek [7]

Answer:

Q5.

a} Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .

b) There excess of Sodium thiosulfate by 2.637 g

Q6.

a) Mass = 162 g Sodium bicarbonate

b)Volume of Ammonia remain after the reaction = 0.086 x 24.8 = 1.60 L

Explanation:

Points to be considered :

There is a difference between STP and SATP :

STP = Standard Temperature and Pressure (273.15 K and 1 atm)

1 mole of gas at STP = 22.4 L

SATP = Standard Ambient Temperature and Pressure (293.15 K and 1 atm)

1 mole of gas at SATP = 24.8 L

$moles = \frac{Given\ mass}{Molar\ mass}$

Number of moles of gas at SATP :

$moles = \frac{Given\ Volume}{24.8L}$

Q5.

First, calculate the number of moles of Cl2 and thiosulfate present in the reaction :

Volume of Cl2 = 105 L

Moles of Cl2 =

$moles = \frac{Given\ Volume}{24.8}$

$moles = \frac{105}{24.8}$

Moles of Cl2 in Reaction medium = 4.2338 mole

Mass of Sodium thiosulfate = 170 g

Molar mass of thiosulfate = 158.11 g/mol (theoretical value)

$moles = \frac{170}{158.11}$

= 1.075

Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole

To check whether the given moles of Cl2 and sodium thiosulfate satisfy theoretical values :

Consider the Given reaction and apply law of conservation of mass

$4Cl_{2}+Na_{2}S_{2}O_{3}+5H_{2}O\rightarrow 2NaHSO_{3}+8HCl$

$Na_{2}S_{2}O_{3}$ = sodium thiosulfate

This equation indicates ,

4 moles of Cl2 require = 1 mole of sodium thiosulfate

1 mole of Cl2 require =

$\frac{1}{4}$ of sodium thiosulfate = 0.25

4.2338 mole of Cl2 should need = 0.25 x 4.2338

= 1.058 mole of sodium thiosulfate

Required Thiosulfate = 1.058 mole

But,

Moles of Sodium Thiosulfate in Reaction medium = 1.075 mole

So , extra moles of Sodium Thiosulfate is present in the reaction by

= 1.075 - 1.0589 = 0.0166 mol

Molar mass of sodium thiosulfate = 158 .11 g/mol

$Mass =0.166\times 158.11$

= 2.637 g

a} .Yes , All Cl2 get eliminated by adding 170 g of sodium thiosulfate .

b) There excess of Sodium thiosulfate by 2.637 g

Q6.

Volume of ammonia = 50.0 L

Moles of Ammonia ,

$moles = \frac{Given\ Volume}{24.8L}$

$moles = \frac{50}{24.8L}$

= 2.016 moles

Moles of CO2 =

Mass of CO2 = 85.0 g

Molar mass = 44 g/mol

$moles = \frac{Given\ mass}{Molar\ mass}$

$moles = \frac{85.0}{44}$

= 1.932 mol of CO2

Now check the law of conservation of mass :

$NaCl + NH_{3} + CO_{2} +H_{2}O\rightarrow NaHCO_{3} +NH_{4}Cl$

According to above equation ,

1 mole of CO2 Needs = 1 mole of NH3

1.93 mol of CO2 need = (1 x 1.93) mol

1.93 mol of CO2 need = 1.93 mol of ammonia

Available ammonia = 2.016 mol

So Ammonia is in excess by:

= 2.016 - 1.93 mol

= 0.086 mol

Volume at SATP is calculated by

$V =mole\times 24.8$

Volume of Ammonia remain after the reaction = 0.086 x 24.8

= 2.1104 L

= 2.10 L

CO2 is the limiting reagent and governs the product formation :

Molar mass of NaHCO3 = 84.007 g/mol

1 mole of CO2 Needs = 1 mole of NaHCO3 = 84.007

1.93 mol of CO2 need = 1.93 x 84.007 mol

= 162.133 g of Sodium bicarbonate

= 162 g Sodium bicarbonate

Note : The answers are present in rounded figures .

6 0

3 years ago