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trasher [3.6K]
3 years ago
14

Which table correctly identifies the subatomic particle's charge and mass? a b c d

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
8 0
It’s b that charts say it
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Calculate the specific heat capacity for a 15.3-g sample of gold that absorbs 87.2 J when its temperature increases from 35.0 °C
diamong [38]

Answer:

The specific heat of gold is 0.129 J/g°C

Explanation:

Step 1: Data given

Mass of gold  = 15.3 grams

Heat absorbed = 87.2 J

Initial temperature = 35.0 °C

Final temperature = 79.2 °C

Step 2:

Q = m*c*ΔT

⇒ Q =the heat absorbed = 87.2 J

⇒ m = the mass of gold = 15.3 grams

⇒ c = the specific heat of gold = TO BE DETERMINED

⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C

87.2 J = 15.3g * c * 44.2°C

c = 87.2 / (15.3 * 44.2)

c = 0.129 J/g°C

The specific heat of gold is 0.129 J/g°C

4 0
3 years ago
Which parts (postulates) of Dalton's atomic theory has been modified in light of later discoveries
professor190 [17]

Explanation:

1) Atoms can not be subdivided. It has been changed as it is possible to divide an atom into Protons, Neutrons and Electrons as well as other smaller particles.

2) The regulations for the chemical mixture have always been altered to the rules for the chemical combination in the creation of organic compounds may be denied.

3)Atoms are of a particular element are identical in all respects i. e. they have same mass and similar properties.

It was modified. And according. to new search;

Atoms of the same element may not be always identical.

8 0
3 years ago
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
3 years ago
What are three ways heat is transfered
Gre4nikov [31]

Answer:

Conduction, Convection and Radiation

Explanation:

6 0
3 years ago
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Is water used to test cheese for protein
nekit [7.7K]

Answer:

yes

Explanation:

because it help the body hydrate

7 0
3 years ago
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