What is the lon # for Beryllium (Be)?

What is the volume of HCl gas required to react with excess magnesium metal to produce 6.82 L of hydrogen gas at 2.19 atm and 35

prohojiy [21]

Answer:

Explanation:

2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)

Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law

P . V = n . R .T

2.19 atm . 6.82L = n . 0.082 . 308K

(2.19 atm . 6.82L) / (0.082 . 308K) = n

0.591 mol = n

1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole

0.591 .2 = 1.182 mole of acid.

Molar mass of HCl = 36.45 g/m

1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g

Density HCl = HCl mass / HCl volume

0,118 g/mL = 43.1 g / HCl volume

43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)

5 0

3 years ago

How many grams of oxygen can be obtained from 250. grams of KCIO3?

guapka [62]

Answer:

Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.

Explanation:

3 0

3 years ago

Which of the following is a reason to make an armature the parent of a creature

REY [17]

you have to show us the rest of it because we have no idea what your looking at. I'm sorry

3 0

3 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

8 0

3 years ago

B. Calculate how many atoms of hydrogen are in 1. 75 moles of glucose (c6h12o6).

ycow [4]

Answer:

1.26x10^25 atoms of hydrogen

Explanation:

because there are 12 atoms of hydrogen in a molecule of glucose, multiply 12 by Avogadro's number (6.02x10^23) to get how many molecules of hydrogen there are in a mole of glucose. Then multiply that number by 1.75, which is the number of moles of glucose there is in this problem.

4 0

3 years ago