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lianna [129]
3 years ago
8

“All compounds containing H atoms are acids, and all compounds containing OH groups are bases.” Do you agree? Give examples.

Chemistry
1 answer:
viktelen [127]3 years ago
3 0

I Do NOT Agree

Explanation:

For these compounds to quality to be acids or basis, they needs to be able to dissociate the H or OH in water for them to be categorized as acids and bases respectively. Compounds that dissociate completely are strong  acids or bases while those that dissociated partially are weak acids or bases.

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22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0
2 years ago
Rank the following compounds from largest to smallest according to their expected magnitude in lattice energy. Rank from largest
fiasKO [112]

Answer:

It's inorder

b,e,d,c and a

Explanation:

Due to electronegativty difference

8 0
3 years ago
Read 2 more answers
A hypothetical element consists of two isotopes of masses 86.95amu and 88.95amu with abundances of 35.5% and 64.5% respectively.
Art [367]

Answer: The average atomic mass of the element = 88.242amu

Explanation:

The abundance of the first isotope is =35.5%

 Atomic mass of first isotope = 68.9257

The average atomic mass of the first isotope =86.95amu X 35.5%  =86.95amu X 0.355 =30.8725 amu

The abundance of the second isotope =64.5%

Atomic mass of the second isotope =88.95amu

The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu

Now the average atomic mass =30.8725 +57.37275 = 88.242amu

OR using the formulae

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100

8,824/100

=88.24amu

5 0
2 years ago
A bulb is found to weigh 5.76 grams more when filled with air at 298 K and 1.00 atmosphere pressure than when it is evacuated. W
garik1379 [7]

Answer:

Subtract water vapor pressure from total pressure to get partial pressure of gas A: PA=1.03 atm- 1 atm=0.03 atm.

What is the total pressure of the gases at 298 K?

98.8 kPa

A sample of nitrogen gas is bubbled through water at 298 K and the volume collected is 250 mL. The total pressure of the gas, which is saturated with water vapour, is found to be 98.8 kPa at 298 K.

The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.

How do you find the partial pressure of water in air?

e is vapor pressure Rv = R∗/Mv = 461.5Jkg−1K−1 and Mv = 18.01gmol−1, ϵ = Mv/Md = 0.622. The vapor pressure is the partial pressure of the water vapor. where es is in Pascals and T is in Celsius.

ExpHow do you find the pressure of h2?

For the high pressures in which hydrogen gas is often stored, the van der Waals equation can be used. It is P+a(n/V)^2=nRT. For diatomic hydrogen gas, a=0.244atm L^2/mol^2 and b=0.0266L/mol.lanation:

6 0
2 years ago
How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 342.6 mL of a 0.783 M LiNO3 solution ? A) 0.00389
Anika [276]

Answer:

b) 18.5 g

Explanation:

The first step is the <u>calculation of number of moles</u> of LiNO_3 with the molarity equation, so:

M=\frac{#~mol}{L}

With the <u>volume in "L" units</u> (342.6 mL= 0.342 L ) we can <u>find the moles</u>, so:

0.783~M\frac{#~mol}{0.342~L}

#~mol=~0.783*0.342=0.268~mol~LiNO_3

Now, we have to calculate the <u>molar mass</u> to convert the moles to grams. For this we have to know the <u>atomic mass</u> of each atom:

Li ( 6.94 g/mol)

N (14 g/mol

O (16 g/mol)

With these values and the number of atoms in the molecule we can do the math:

(6.94 x 1) + ( 14 x 1) + (16 x 3 ) = 68.94 g/mol

The final step is the <u>calculation of the grams</u>, so:

0.268~mol~LiNO_3~\frac{69.94~g~LiNO_3}{1~mol~LiNO_3}=~18.5~g~LiNO_3

<u>We will need 18.5 g of LiNO3 to do a 0.783 M solution with a volume 342.6 mL.</u>

4 0
3 years ago
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