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ki77a [65]
3 years ago
7

Which process is used to produce gases from solutions of salts dissolved in water or another liquid?

Chemistry
2 answers:
icang [17]3 years ago
5 0

Answer:

D.electrolysis

Explanation:

sana makatulong

NemiM [27]3 years ago
3 0

Answer:

Explanation:

Option D electrolysis is the correct answer

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93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP
Dahasolnce [82]
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
3 0
3 years ago
Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas?Ne, N₂,
malfutka [58]

Answer:

Ne

Explanation:

Atomic number of Ne is 10.

Electronic configuration of Ne:

1s^2 2s^2 2p^6

Octet of Ne is complete . Element having complete octet are stable and behave ideal gas.

N_2 and CH_4 are reactive and hence, does not behave as ideal gas.

3 0
3 years ago
Please help with number 24
vladimir1956 [14]
Layla it is A. that's the only one in standard form.
3 0
3 years ago
Read 2 more answers
How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0
3 years ago
2.75 x 4 x 10000 (with explanation)
Murljashka [212]

Answer:

110000

Explanation:

Multiply 2.75 by 4. is 11

Multiply 11 by 10000. is

110000

6 0
1 year ago
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