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Eva8 [605]
3 years ago
13

Giving brainliest if correct!! please help someone out?

Chemistry
1 answer:
Stels [109]3 years ago
5 0

Answer:

protons: 36

neutrons:48

electrons:36

Explanation:

the number of protons in an element is = to the atomic # (36)

the number of neutrons is the atomic mass - atomic # (84-36=48)

In a neutral charged element the # of protons = # of electrons

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A 30.5 g sample of an alloy at 95.0°C is placed into 49.3 g water at 24.3°C in an insulated coffee cup. The heat capacity of the
user100 [1]

Answer:

0.752 J/g*K

Explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

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State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl
Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

\Delta S = S(products)-S(reactants)----(1)

a) PCl3(l) + Cl2(g) \rightarrow PCl5(s)

Here  the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) 2HgO(s) \rightarrow 2Hg(l) + O2(g)

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) H2(g) \rightarrow 2H(g)

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) U(s) + 3F2(g) \rightarrow UF6(s)

Here  one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the  product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0
3 years ago
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