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Leni [432]
3 years ago
11

. A standard dry cell has an output voltage of A. 1.5 VDC. B. 1.1 VDC. C. 1.2 VDC. D. 2.0 VDC.

Chemistry
1 answer:
NikAS [45]3 years ago
4 0
<span>The correct answer is letter A. 1.5 VDC. A standard dry cell has an output voltage of A. 1.5 VDC. Standard dry cell is a type of electricity-producing chemical cell, that is commonly use in households, and even portable devices. Dry cell is zinc-carbon cell and with nominal voltage of 1.5 volts.</span>
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How do you find the pressure in a container if you don't have the temperature? 
BARSIC [14]
It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
PV = constant or P1V1 = P2V2 where P1V1 are initial conditions and P2V2 are final.


For adiabatic process,

PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.

where gamma = Cp
                          ------
                           Cv
Cp = specific heat at constant pressure and Cv = specific at constant volume.

Value of Gamma will be given in question.
Hope this helps!


7 0
3 years ago
Please help me this is a big grade please
Readme [11.4K]

Answer:

water

Explanation:

because the others are in compounds and can not be considered as  products lonely.

5 0
2 years ago
Which is an example of diffraction?
jeyben [28]
I would say d but I’m not sure
8 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
Does percent error measure accuracy or precision?
Nesterboy [21]

Answer:

Precision

Explanation:

It figures out how close all of the data numbers are.

4 0
3 years ago
Read 2 more answers
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