Do you think precipitation has a greater impact on the rate of chemical weathering or mechanical weathering? Explain

1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.

svet-max [94.6K]

Answer:

(a) Theoretical Yield = 10.50 g

(b) %age yield = 27.33 %

Explanation:

Answer-Part-(a)

The balance chemical equation for the synthesis of Ammonia is as follow;

N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

Moles = Mass / M/Mass

Moles = 9.75 g / 28.01 g/mol

Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

Moles = Mass / M/Mass

Moles = 1.86 g / 2.01 g/mol

Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

1 mole of N₂ reacts with = 3 moles of H₂

So,

0.348 moles of N₂ will react with = X moles of H₂

Solving for X,

X = 3 mol × 0.348 mol / 1 mol

X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

3 mole of H₂ produces = 2 moles of NH₃

So,

0.925 moles of H₂ will produce = X moles of NH₃

Solving for X,

X = 2 mol × 0.925 mol / 3 mol

X = 0.616 mol of NH₃

Step 5: Calculating theoretical yield of Ammonia as,

Theoretical Yield = Moles × M.Mass

Theoretical Yield = 0.616 mol × 17.03 g/mol

Theoretical Yield = 10.50 g

Answer-Part-(b)

%age yield = Actual Yield / Theoretical Yield × 100

%age yield = 2.87 g / 10.50 g × 100

%age yield = 27.33 %

4 0

3 years ago

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

6 0

3 years ago

What is the effect of adding more water to the following reaction at equilibrium? CO2 + H2O H.CO.

LenaWriter [7]

Answer:If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change. If the pressure is increased the equilibrium will shift to favour a decrease in pressure.

Explanation:

Slidin brainliest??

3 0

3 years ago

How many moles of electrons are required to reduce one mole of nitrogen gas (N2) to two moles of nitrogen ions (N3-)?

Makovka662 [10]

Start by writing the atoms balance:

$N_{2} -\ \textgreater \ 2N^{3-}$

Now, determine the change of oxidation states: $N_{2}$ has oxidation state 0, so each N has to gain 3 electrons to become $N^{3-}$ .

That, means that you need 6 electrons to balance the charges, resulting in:

$N_{2} + 6 e^{-} -\ \textgreater \ 2 N_{3-}$

And the answer is 6 mole of electrons.

8 0

3 years ago

What is the ratio strength of 100 ml containing a 1:50 povidone iodine solution diluted to 1000 ml? (answer must be numeric; no

guapka [62]

We are given that:

100 mL total solution with 1:50 povidone iodine solution

This means that there is 1 mL of povidone iodine solution per 50 mL of total solution. Since we are given a total of 100 mL solution, therefore we have an initial amount of:

pure povidone iodine solution = 2 mL

This amount of pure povidone iodine solution is added or diluted (most perhaps with water) to make a total of 1000 mL total solution, therefore the new ratio is:

2:1000 povidone iodine solution

By dividing both sides by 2, this simplifies to

1:500 povidone iodine solution

Answer:

1:500

4 0

3 years ago