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mestny [16]
3 years ago
8

nter your answer in the provided box. Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. C

alculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.39 M. pH
Chemistry
1 answer:
OLga [1]3 years ago
8 0

Answer:

Explanation:

pH = - log[ H⁺ ]

[ H⁺ ] = .39

pH = - log .39

= - log (39 x 10⁻²)

= 2 - log 39

= 2 - 1.59

=  .4 1

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Which of the following substances is never a Brønsted-Lowry base in an aqueous solution?Group of answer choicespotassium hydroxi
erik [133]

i dont know because im stupid

4 0
3 years ago
Where is the mass of an atom found? Explain.
muminat

Explanation:

Over 99.9% percent of an atom's mass resides in the nucleus. The protons and neutrons in the center of the atom are about 2000 times heavier than the electrons orbiting around it.

Because the electrons are so light by comparison,they represent only a tiny fraction of a percent of the atom's total weight.

Hope this answer helps you!

4 0
3 years ago
A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen
hammer [34]

Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

U_1=U_2

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: U_1=U_1^C+U_1^{N_2}

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}

So, the overall energy balance is:

U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}

Reorganizing,

U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}

Take in mind that, for the mass balance for nitrogen, m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2},

So, let's replace m_1^{N_2} in the energy balance:

m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

So, as you can see, the term m_{2,liq}^{N_2}u_{2,liq}^{N_2} disappear because u_{2,liq}^{N_2}=u_{1,liq}^{N_2} (The specific energy in the liquid is the same because the temperature does not change).

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}

m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})

The difference (u_{2,vap}^{N_2}-u_1^{N_2}) is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}

m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg

3 0
4 years ago
i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t
Ronch [10]
Assuming that the number of mols are constant for both conditions:
\frac{P_1V_1}{T_1} =  \frac{P_2V_2}{T_2}
Now you plug in the given values. V_1 is the unknown. 
\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}

Separate V_1
V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }
V= 162.782608696 L 

There are 2 sig figs

V= 160 L
3 0
3 years ago
Match the atomic particles with their electrical charges.
PilotLPTM [1.2K]

Answer:

proton: positive

electron: negative

neutron: neutral

Looks like you already have it.

7 0
3 years ago
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