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shusha [124]
3 years ago
8

Are pure substances homogeneous or heterogeneous

Chemistry
1 answer:
Svetllana [295]3 years ago
6 0

The answer is homogenous i think

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Uso industrial hidruro mercurico
bonufazy [111]

he arsenic acid or arsenate hydrogen as it is also known to this compound (H 3 AsO 4 ) is the acid form of <span>ion </span>arsenate , AsO<span>4 </span>3- , one anion trivalent in which arsenic has an oxidation state of + 5. Chemically, arsenates behave in a similar way tophosphates .

There is another compound derived from this one that is the arsenious acid or arsenite of hydrogen

5 0
3 years ago
Identify a cation. An atom that has gained a proton. An atom that has lost an electron. An atom that has gained an electron. An
iren2701 [21]

the answer would be B an atom that has lost an electron

8 0
2 years ago
What is the charge for the compound Cr(OH)3
devlian [24]
The answer is an answer
8 0
3 years ago
What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?
kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of Al(OH)_3 = 0.610 g

Molar mass of Al(OH)_3 = 78 g/mol

\text{Number of moles}=\frac{0.610g}{78g/mol}

Number of moles of Al(OH)_3 = 0.0078 moles

The reaction between Al(OH)_3 and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

By Stoichiometry,

1 mole of  Al(OH)_3 reacts with 3 moles of HCl

So, 0.0078 moles of Al(OH)_3 will react with \frac{3}{1}\times 0.0078 = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

0.0234moles=\frac{\text{Given mass}}{36.5g/mol}

Mass of HCl required will be = 0.8541 grams

3 0
3 years ago
Normally the capital ÎGo° for a reaction would be determined at standard temperature with each reactant at a concentration of 1
Arte-miy333 [17]

Answer:

d.-379 cal/mol

Explanation:

ΔG = ΔG⁰ + RT ln K

for equilibrium ΔG = 0

ΔG⁰ + RT ln K =0

ΔG⁰   =  -  RT ln K

PG ⇒ PEP

K = [ PEP ] / [ PG ]

= .68 / .32

= 2.125

ΔG⁰   =   - 1.987 x 273 x  ln 2.125

= - 409 Cal / mole

Option d is the nearest answer .

8 0
3 years ago
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