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Ahat [919]
3 years ago
6

a 55g block of metal has a specific heat of 0.45 J/g°C. What will be the temperature change of this metal if 450 J of heat energ

y are added
Chemistry
1 answer:
Dafna11 [192]3 years ago
7 0

Answer:

18.18C

Explanation: Write whats below to show your work :D

m=55

C=0.45

Q=450zj

ch.temp=?

Q=mct

450 = (55)(.45)T

450 = 24.75

/24.75 = /24.75 (cancels out)

18.18 is the answer

Hope this helps! did it with my teacher.

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Which of the following is the correct measurement for the volume of liquid shown below?
mr Goodwill [35]

Answer:

A. 3.5ml

Explanation:

lowest point of liquid or meniscus

5 0
3 years ago
There are 0.3 moles of xenon gas in a 0.5-liter container at 30 degrees C. What is the pressure exerted by the xenon gas
Kaylis [27]
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
P = 14.9258934 atm
3 0
3 years ago
Read 2 more answers
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
4 years ago
Thank uuuu so much!!!!!
kow [346]

Answer:

A. 0.83

Explanation:

Hopefully this helps!

7 0
3 years ago
PLEASE HURRY!! WILL GIVE BRAINLIEST!!!!
Snowcat [4.5K]

Answer:

a

Explanation:

8 0
3 years ago
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