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alexdok [17]
3 years ago
12

2. Pietro decided to travel along the path shown below. How far south did Pietro travel?(The figure

Chemistry
1 answer:
kotegsom [21]3 years ago
3 0
Where’s the path? can’t answer the question without seeing the path so if you insert it I will be able to answer it
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Identify the location of each element group on the periodic table. ​
Gwar [14]
What? sorry i’m confused
8 0
3 years ago
Read 2 more answers
Why do you think haiti <br> is poor or not
Shalnov [3]

Answer:

It kinda is

Explanation:

The school system isn't very affective on the young adolcent minds, the crime rate is very high (indicating poverty rates) some drug smmuglers are based in Haiti, and half the city is destroyed from either people or the environment.

6 0
2 years ago
The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
3 years ago
I was checking my answers for a chem test and came across a solution for a problem where the ∆S was pos, ∆H was neg and ∆G was n
nevsk [136]

Answer:

there is only 15 points

Explanation:

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: \Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0ΔS  

universe

​

=ΔS  

system

​

+ΔS  

surroundings

​

>0delta, start text, S, end text, start subscript, start text, u, n, i, v, e, r, s, e, end text, end subscript, equals, delta, start text, S, end text, start subscript, start text, s, y, s, t, e, m, end text, end subscript, plus, delta, start text, S, end text, start subscript, start text, s, u, r, r, o, u, n, d, i, n, g, s, end text, end subscript, is greater than, 0

At constant temperature and pressure, the change in Gibbs free energy is defined as \Delta \text G = \Delta \text H - \text{T}\Delta \text SΔG=ΔH−TΔSdelta, start text, G, end text, equals, delta, start text, H, end text, minus, start text, T, end text, delta, start text, S, end text.

When \Delta \text GΔGdelta, start text, G, end text is negative, a process will proceed spontaneously and is referred to as exergonic.

The spontaneity of a process can depend on the temperature.

Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:

8 0
3 years ago
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective b
Sonja [21]

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

The reaction product is a single solid and the  the reactants were 2 species in solution. The change in entropy is negative.

CaCO3(s)→CaO(s)+CO2(g)2

Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

We have 4 mole gases as products starting from 1 mol reactant gas, the entropy has increased.

2Na(s)+Cl2(g)→2NaCl(s)

In this reaction 2 mol solid Na and 1 mol Cl₂ gas are converted into 2 mol solid NaCl, the entropy has decreased.

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.

4 0
3 years ago
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