Which one of these equations is balanced? a. Zn + 2hcl --> ZnCl2 + H2 b. S8 + 24F2 --> 8SF6

Lisa [10]

Answer:

Explanation: S + 6 HNO3 --> H2SO4 + 6 NO2 + 2 H2O In the above equation how many moles of water can be made when 115.3 grams of HNO3 are consumed? Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0 Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.: Element Molar Mass Hydrogen 1 Nitrogen 14 Sulfur 32 Oxygen 16...

Hope it helps..!!

4 0

2 years ago

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Unstable isotopes occur when the strong force is unable to overcome the _________.

Alecsey [184]

Unstable isotopes occur when the strong force is unable to overcome the electrostatic force.
There are no stable isotopes in the elements at the upper end of the periodic table, which clearly demonstrates the limit of the ability of the nuclear binding energy or the residual strong force, to overcome the electrostatic repulsion of all those protons in the nucleus.

6 0

2 years ago

Would you expect the substance to float or sink in water if mass is 25g and volume is 7 cm

Sonbull [250]

Answer:

Sink

Explanation:

For a substance to float in water, it must be less dense than water. The density of water is 1 g/cm^3. To find the density of the substance, we can do mass/volume, which in this case is 25g/7cm^3. We see that $25/7 > 1$ , so the substance would sink.

4 0

1 year ago

1. Draw a diagram to show how energy is transferred by water particles as a wave moves through water

castortr0y [4]

Answer:

uh can search on Google dear

7 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

2 years ago