Given the formula 4f10, what does the 10 stand for?

AgNO3(aq)+Cu(s)→Ag(s)+Cu(NO3)2(aq) is an example of which type of reaction?

Elan Coil [88]

Its a displacement reaction

7 0

4 years ago

What is the relationship between mole, Avogadro number and mass?

weeeeeb [17]

Answer:

The mass of one mole of a substance is equal to that substance's molecular weight. ... water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams. ... Avogadro's number is a proportion that relates molar mass on an atomic ... one molecule of water (H2O), one mole of oxygen (6.022×1023 of O atoms)

5 0

3 years ago

A closed container has 5.60 x 1023 molecules of oxygen gas (O2). How many moles is that?

ozzi

The answer is 5.60 x 1023= 5,728.8

8 0

3 years ago

In a chemical change energy can be ___?

qaws [65]

Answer:

chemical change energy can neither be created nor destroyed.

Explanation:

the rule of law of conversion states that energy can only be converted from said energy to another. it cannot be created and destroyed.

7 0

3 years ago

Read 2 more answers

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5

victus00 [196]

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

where ;

$P_1$ is the vapor pressure at temperature 1

$P_ 2$ is the vapor pressure at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

$P_1$ = 1 atm

$P_ 2$ = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

$T_1$ = 282 °C = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values into the Clausius - Clapeyron equation, we have:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }$

$\dfrac{1}{T_2} = 0.002108048708$

$T_2 = \dfrac{1}{0.002108048708}$

$\mathbf{T_2 }$ = 474.37 K

To °C ; we have $\mathbf{T_2 }$ = (474.37 - 273)°C

$\mathbf{T_2 }$ = 201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

3 0

3 years ago