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3 years ago

Can someone please help me calculate the pH and pOH of these?????

Chemistry

1 answer:

FinnZ [79.3K]3 years ago

4 0

a. pH=5, pOH=9

b. pH=6.4771, pOH=7.5229

c. pH=1.6021, pOH=12.3979

d. pH=11.8751, pOH=2.1249

e. pH=13.9208, pOH=0.0792

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22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp

IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

Moles NH3 -Molar mass: -17.01g/mol-

Using PV = nRT

PV/RT = n

Where P is pressure = 1atm at STP

V is volume = 22.4L

R is gas constant = 0.082atmL/molK

T is absolute temperature = 273.15K

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, the limiting reactant is NH3

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

4 0

2 years ago

Which of the following chemical formulas represents a common acid?

wariber [46]

HCI is one of the most common acids out of the following

5 0

3 years ago

Number of electrons for all of these even the ones with a number by them ? Will give brainlist!!!!

Bumek [7]

Explanation:

s=2

p=6

d=10

f=14

pls mark me brainlist

7 0

2 years ago

I’m very confused on how to solve this

Bond [772]

Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place

4 0

3 years ago

3. Given 20g of Barium Hydroxide, how many grams of

anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present = $\frac{20}{171.34}$

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

6 0

2 years ago