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kherson [118]
3 years ago
12

For maleic acid, hoocch=chcooh, ka1 = 1.42  10–2 and ka2 = 8.57  10–7 . what is the concentration of maleate ion (–oocch=chcoo

– ) in a 0.150 m aqueous solution of maleic acid?

Chemistry
2 answers:
uranmaximum [27]3 years ago
7 0

ANSWER

Find the attachment

Talja [164]3 years ago
6 0

Answer:

Explanation:

solution to the question

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If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.
Aliun [14]

Answer:

1 L

Explanation:

ppm means parts per million. Generally the relationship between mass and litre is given as;

1 ppm = 1 mg/L

This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.

3 0
4 years ago
. Given the reaction 2HgO → 2Hg + O2 , how many moles of elemental mercury will be obtained by the decomposition of 1 mole of Hg
AveGali [126]

Answer:

one mole of HgO will give one mole of Hg.

Explanation:

Given data:

Moles of HgO = 1 mol

Moles of Hg = ?

Solution:

Chemical equation:

2HgO → 2Hg + O₂

Now we will compare the moles of Hg with HgO from balance chemical equation.

                  HgO   :     Hg

                    2       :      2

                     1       :      2/2×1 = 1 mol

So, one mole of HgO will give one mole of Hg.

6 0
3 years ago
Light-dependent reactions: ​
dmitriy555 [2]
Ya makes totally sense
3 0
3 years ago
What mass of dinitrogen monoxide, N2O, contains the same number of molecules as 3.00 g of trichlorofluoromethane, CCl3F?
erica [24]

Answer:

0.9612 g

Explanation:

First we <u>calculate how many moles are there in 3.00 g of CCl₃F</u>, using its <em>molar mass</em>:

  • 3.00 g CCl₃F ÷ 137.37 g/mol = 0.0218 mol CCl₃F

Now, we need to calculate how many grams of N₂O would have that same number of molecules, or in other words, <em>the same amount of moles</em>.

Thus we <u>calculate how many grams would 0.0218 moles of N₂O weigh</u>, using the <em>molar mass of N₂O</em> :

  • 0.0218 mol N₂O * 44.013 g/mol = 0.9612 g N₂O
3 0
3 years ago
A 12.82 g sample of a compound contains 4.09 g potassium (K), 3.71 g chlorine (Cl), and oxygen (O). Calculate the empirical form
Rina8888 [55]
The  empirical  formula of the  compound is  calculated  as  follows

first   calculate  the  mass  of  oxygen=  12-(4.09  +3.71)=  5.02g

then  calculate  the  moles  of  each  element,  moles  =  mass/  molar  mass

moles of   K  =  4.09g/39 g/mol(molar  mass  of K)  =  0.105  moles
moles  of Cl = 3.71g/35.5 g/mol(molar  mass  of Cl) =  0.105  moles
moles of  O  =  5.02g/ 16g/mol(molar  mass of  O) = 0.314  moles

then  calculate e  mole ratio by  dividing  each  mole  by  the  smallest  number  of  moles  (  0.105 moles)

K=0.105/0.105= 1
Cl=0.105 /0.105=1
O=  0.314/0.105=3

therefore  the  empirical   formula  = KClO3
7 0
3 years ago
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