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s2008m [1.1K]
4 years ago
15

1. _ LiBr2 + __K2CO3 -->___ LiCO3 +_ KBr Reaction Type:

Chemistry
1 answer:
Arada [10]4 years ago
5 0

Answer:

Double displacement chemical reaction.

Explanation:

Hello,

In this case, for the given reaction we firstly balance it:

LiBr_2 +K_2CO3 \rightarrow LiCO3 +2KBr

Then, since the lithium and potassium cations are being exchanged to each other from bromide and carbonate, we are talking about a double displacement chemical reaction.

Best regards.

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What kind of data would you need to collect to carry out this experiment?
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A gas mixture being used to simulate the atmosphere of another planet at 23°c consists of 337 mg of methane, 148 mg of argon, an
Karolina [17]

The total pressure of the mixture is 65.5 kPa.

According to Dalton's Law of Partial Pressure,

The partial pressure of gas = Mole fraction of gas × Total pressure

Total Pressure = Sum of all the gases partial pressures

The number of moles of methane is,

Moles \:  of \: methane  \: (16 g/mol) =  337 \: mg  \times  \frac{1 g}{1000 mg} \times  \frac{ 1 mol}{16 g }

= 0.021 mols

The moles of methane are 0.021 mols.

The number of moles of the argon,

Moles \:  of \: argon (40 g/mol) = 148 \:  mg  \times  \frac{  1 g}{1000 mg } \times  \frac{  1 mol}{40 g}

= 0.003 mols

The number of moles of argon is 0.003 mols.

The number of moles of nitrogen is,

Moles  \: of \: nitrogen (28 g/mol) = 296 \:  mg  \times  \frac{ 1 g}{1000 mg}  \times  \frac{  1 mol/}{28 g}

= 0.010 mols

The number of moles of nitrogen is 0.010 mols.

The total number of moles is,

= 0.021 + 0.003 + 0.010

= 0.034 mols

Mole \:  fraction =  \frac{ Moles \:  of \:  solute }{Total \:  number  \: of  \:  moles  \: of  \: soulte \:  and \:  solvent}

= \frac{  0.010 }{ 0.034}

= 0.29

0.29 \: P _{total} = 19 \:  kPa

P _{total} =  \frac{ 19  \: kPa }{0.29}

= 65.5 kPa

Therefore, the total pressure of the mixture is 65.5 kPa.

To know more about Dalton's law, refer to the below link:

brainly.com/question/14119417

#SPJ4

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1 year ago
Changes in appearance...
Serjik [45]
I'm not quite sure but I think if you field the corner or cut it it would physically change it while not creating something new (chemical change).
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4 years ago
Assume that concentrated aqueous NH3 has a density of 0.252 g/mL (0.252 g of NH3 per mL of liquid). Calculate the volume of NH3
Kobotan [32]

The question is incomplete, here is the complete question:

Assume that concentrated aqueous NH₃ has a density of 0.252 g/mL (0.252 g of NH₃ per mL of liquid). Calculate the volume of NH₃ required to contain 0.442 mol?

<u>Answer:</u> The volume of ammonia required is 29.82 mL

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.442 moles

Putting values in above equation, we get:

0.442mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(0.442mol\times 17g/mol)=7.514g

To calculate the volume of ammonia, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ammonia = 0.252 g/mL

Mass of ammonia = 7.514 g

Putting values in above equation, we get:

0.252g/mL=\frac{7.514g}{\text{Volume of ammonia}}\\\\\text{Volume of ammonia}=\frac{7.514g}{0.252g/mL}=29.82mL

Hence, the volume of ammonia required is 29.82 mL

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How many atoms are represented in 28 grand of sodium
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