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ExtremeBDS [4]
2 years ago
8

Please answer 14 will give brainliest

Chemistry
2 answers:
Setler [38]2 years ago
7 0

Answer:

its \: fluorine \: and \: it \: belongs \: to \ :  \\ 17th \: family \: noble \: gases

rosijanka [135]2 years ago
4 0

Answer:

I think it's alkali metals

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a sample of substance x that has a mass of 326.0 g releases 4325.8 cal when it freezes at its freezing point. if substance x has
ira [324]
1) Calculate the number of mols,n,  of the substance

n = mass/ molar mass = 326.0 g / 58.45 g / mol = 5.577 moles

2) Calculate the molar heat of fusion as the total heat released by the sample divided by the number of moles

hf = heat released / n = 4325.8 cal / 5.577 moles = 775. 59 cal /mol
8 0
3 years ago
Read 2 more answers
The following questions pertain to a system contains 122 g CO(g) in a 0.400 L container at -71.2 degrees C.
7nadin3 [17]

Answer:

a. P = 182 atm

Explanation:

Data Given:

amount of CO = 122g

Volume of CO = .400 L

Temperature of CO =  -71.2°C

Convert the temperature to Kelvin

T = °C + 273

T =  -71.2 + 273

T =  201.8 K

a. Calculate the pressure exerted by the CO(g) in this system using the ideal gas equation (P) = ?

Solution:

To calculate Pressure by using ideal gas formula

                  PV = nRT

Rearrange the equation for Pressure

                   P = nRT / V . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant = 0.08206 L.atm / mol. K

For this we have to know the mole of the gas and the following formula will be used

                 no. of moles = mass in grams / molar mass . . . . . . (2)

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in equation 2

                no. of moles = 122 g / 28 g/mol

                no. of moles = 4.4 mol

Now put the value in formula (1) to calculate Pressure for CO

P = 4.4 x 201.8 K x 0.08206 (L.atm/mol. K) / 0.400 L

P = 182 atm

So the pressure will be 182 atm

__________

b. Data Given:

Actual pressure exerted by CO = 145 atm

expected pressure exerted by CO = 182 atm

why the actual pressure is less than what would be expected = ?

Explanation:

This is because of the deviation from ideal behavior of real gases.

The real gases approach to ideal behavior under very high temperature and very low pressure.

But CO deviate from ideal behavior to give expected value for pressure, because it behave at high pressure and low temperature.

This non-ideal behavior is due to two postulate of ideal behavior

  • gas molecules have negligible volume
  • Gas molecules have negligible inter-molecular interaction

but these postulates not obeyed under real condition. so we calculated the pressure using ideal condition values for gas and obtained the expected value for pressure but the actual pressure value was detected under normal condition.

8 0
3 years ago
In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is
Verizon [17]

Answer:

a) 10 mg NO_3^-/L

b) 1.61*10^{-4}mol NO_3^-/L

c) 2.26 mg N/L

Carbon: C=26.64 \frac{mg C}{L}

Explanation:

<u>Nitrate</u>

First of all, is important to know that:

1 ppm=1 mg/L

a) 10 ppm of nitrate (NO_3^-) is equal to 10 mg NO_3^-/L

b) The molecular weight of nitrate is 62 g NO_3^-/mol

10 mg NO_3^-/L=0.01 g NO_3^-/L

\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L

c) Nitrate has 14 mg of N per 62 mg of NO3

10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L

<u>Carbon</u>

Carbonate has 12 mg of C per 60 mg of CO_3^{-2}

Bicarbonate has 12 mg of C per 61 mg of HCO_3^{-}

C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}

C=26.64 \frac{mg C}{L}

8 0
3 years ago
Suppose that 10.0 L of Carbon Dioxide gas are produced by this reaction, 4C3H5N3O9 -&gt; 12 CO2 + 10H2O + 6N2 +O2, at a temperat
siniylev [52]

The mass of nitroglycerin : 34.52 g

<h3>Further explanation</h3>

Reaction

4C₃H₅N₃O₉ ⇒ 12 CO₂ + 10H₂O + 6N₂ +O₂

Volume = 10 L

Temperature = -5°C=268 °K

Pressure = 1 atm

mol of CO₂ (ideal gas) :

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 10}{0.082\times 268}\\\\n=0.455

mol ratio C₃H₅N₃O₉ : mol CO₂= 4 : 12, so mol C₃H₅N₃O₉ :

\tt \dfrac{4}{12}\times 0.455=0.152

mass C₃H₅N₃O₉ (MW=227,0865 g/mol):

\tt 0.152\times 227.0865=\boxed{\bold{34.52~g}}

4 0
3 years ago
We know that for a given reaction when the temperature increases from 100 k to 200 k the rate constant doubles. What is the acti
Alenkasestr [34]

Answer:

The activation energy of the reaction is 1.152 kJ/mol.

Explanation:

Activation energy is the minimum amount which is absorbed by the reactant molecules to undergo chemical reaction.

Initial temperature of reaction = T_1=100 K

Final temperature of reaction = T_2=200 K

Initial rate of the reaction at 100 k = K_1=k

Final rate of the reaction at 200 k = K_2=2k

Activation energy is calculated from the formula:

\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}(\frac{T_2-T_1}{T_1T_2})

R = Universal gas constant = 8.314 J/ K mol

\log\frac{2k}{k}=\frac{E_a}{2.303\times 8.314 J/K mol}(\frac{200 K-100K}{200 K\times 100K})

E_a=1,152.772 J/mol=1.152 kJ/mol

5 0
3 years ago
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