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krok68 [10]
3 years ago
13

Solutions that are very concentrated have greater freezing point depression Group of answer choices true or false

Chemistry
1 answer:
Tamiku [17]3 years ago
8 0

Answer:

Explanation:

False. The greater the concentration, the lower the freezing point.

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Gas Law- please help!!!
jekas [21]

The pressure of the carbon dioxide will be 0.09079 atm.

<h3>What is partial pressure?</h3>

The pressure exerted by the individual gas is known as partial pressure.

The partial pressure is given as

\rm P_{Total} = P_1 + P_2 + ...

In a mixture of carbon dioxide and oxygen, 40.0% of the gas pressure is exerted by oxygen.

If the total pressure is 115 mmHg.

The total pressure in atm will be

P = 115 mmHg

P = 0.15132 atm

We have

\rm P_T \ \  \  = P_{O_2} + P_{CO_2}\\\\0.15132 \ \  = 0.4 \times 0.15132 +P_{CO_2}\\\\ P_{CO_2} = 0.09079\ atm

Then the pressure of the carbon dioxide will be 0.09079 atm.

More about the partial pressure link is given below.

brainly.com/question/13199169

#SPJ1

8 0
2 years ago
When an electron falls from an excited state to ground state, is energy released or absorbed?
strojnjashka [21]

Answer:

Energy is released.

Explanation:

When an electron absorbs energy, it moves up into an excited state. When it releases energy, it will return to the ground state.

5 0
3 years ago
After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid
nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\

Then, we add the half reactions:

2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0
2 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
Selected Answer: alpha particle
Lostsunrise [7]

Answer:

positron

Explanation:

A positron is a particle produced when a proton is transformed into a neutron. Anti neutrinos are ejected from the nucleus to balance spins.

Positron emission increases the Neutron/Proton ratio. When a nuclide undergoes positron emission, the atomic number of the daughter nucleus is one unit less than that of the parent nucleus hence it is found one place before its parent in the periodic table.

3 0
3 years ago
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