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weeeeeb [17]
2 years ago
5

Which of the following correctly compares glass A and glass B?

Chemistry
2 answers:
Sidana [21]2 years ago
8 0

Answer:

Glass A has a greater mass than glass B

Vlad1618 [11]2 years ago
5 0

Answer:

Glass A has a greater mass than Glass B

hope this helped :>

You might be interested in
How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha
VLD [36.1K]
This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(<span>333,550 J/kg)
</span>ΔQ = 5,503,780 J

5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
<em>Δt = 122,434.691 s or 34 hours</em>

4 0
3 years ago
I want to write the answer down so please answer it step by step
olga55 [171]
I suck at chemistry but i have a friend that can help
6 0
3 years ago
Which of the following is a nonrenewable resource?
Studentka2010 [4]

The answer is C) Nuclear.

The answer is C because, Renewable resources are something that can be re-used over and over again. Then Nonrenewable resources can't be made right away once it's been used. That being said, the answer is C.

A) Geothermal can be reused, which makes it renewable.

B) Solar can also be reused, which makes it renewable.

C) Nuclear can NOT be reused, which makes it nonrenewable.

D) Biomas is indeed renewable.

8 0
3 years ago
B. If the sand you ran across has a specific-heat capacity of 835 J/(kgºc),
Murrr4er [49]

Answer: 16700 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed = ?

m = mass of sand = 2 kg

c = heat capacity = 835J/kg^0C

Initial temperature  = T_i = 40^0C

Final temperature= T_f  = 50^0C

Change in temperature ,\Delta T=T_f-T_i=(50-40)^0C=10^0C

Putting in the values, we get:

Q=2kg\times 835J/kg^0C\times 10^0C

Q=16700J

16700 J of energy must be added to a 2-kilogram pile of it to increase its temperature from 40°C to 50°C

4 0
3 years ago
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
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