1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrei [34K]
3 years ago
13

Describe what a geologist would observe if she poured acid on the sculpture

Chemistry
1 answer:
natita [175]3 years ago
6 0

Answer:

A geologist would observe how the material that the sculpture is made out of reacts to the acid.

You might be interested in
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of
snow_tiger [21]

Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.

8 0
3 years ago
Substance A has the following properties.
givi [52]

A curve of temperature vs. time for the entire heating process.

The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.

i ) Calculating the heat and time required to heat the sample to its boiling point:

Boiling point = 85°C

C(liquid) = 2.5 J/g °C

The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,

Therefore, T f = 85°C  and T i = - 20°C

Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,

q3 = 25 g ×  2.5 J/g °C × [85 - (-20)]°C

     = 25 J/°C ×[85+20]°C

     = 6562.5 J

The total heat energy required for heating the sample from initial temperature to boiling point is:-

q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J

                    = 11562.5 J

The Rate of heating = 450 J/min

450. J = 1 min

   11562.5 J = ? min

11562.5 J × 1min/450 J = 25.69 min

ii) Calculating the heat and time required to boil the sample:

∆H Vap = 500 J/g

The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined

q4= m × ∆Hvap

    = 25 g × 500 J/g

   = 12500 J

Thus, total heat required to this phase change is q1 + q2 + q3 + q4  = 500 J + 4500 J +6562.5  J + 12500 J = 24062.5 J

The Rate of heating = 450 J / min

450 J = 1 min

24062.5 J = ? min

24062.5J ×  1min / 450 J = 53.47 min

iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C

C gas = 0.5 J / g °C

The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,

Therefore, T f = 100.°C  and T i = 85°C

q5 = 25 g ×  0.5 J / g °C × [100 - 85] °C

    = 25 J / °C ×15 °C

    = 187.5  J

The total heat energy required for heating the sample from initial temperature to 100°C is

q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J

                                      =24250 J

The Rate of heating = 450 J / min

  450. J = 1 min

 24250 J=? min

Thus, heating the sample to 100.°C takes a total of 53.89 min.

iv) Draw a curve of temperature vs. time for the entire heating process:-

Temperature °C     Temperature K     Heat energy (J)     Time (min)

 -40 °C                       233                             0                     0

-20 °C                          253                          500                  1.11    

Melting -20 °C             253                        5000                   11.11

85 °C                         358                         11562.5              25.69

Boiling 85 °C             358                           24062.5          53.475              

100  °C                       373                             24250          53.89

Hence, the graph for the result is in the image.

Learn more about temperature here:-brainly.com/question/24746268

#SPJ4

4 0
2 years ago
HNO3 + H2SO4 + NO
77julia77 [94]

Explanation:

The two half equations are;

3e + HNO3 → NO

S→ H2SO4 + 6e

When balancing half equations, we have to make sure the number of electrons gained is equal to the number of electrons lost.

<em>Which factor will you use for the top equation?</em>

We multiply by 2 to make the number of electrons =  6e

<em>Which factor will you use for the bottom equation?</em>

We multiply by 1 to make the number of electrons = 6e

3 0
3 years ago
Silicon Tetrachloride is reacted with very pure magnesium, producing silicon and magnesium chloride. (What would be the unbalanc
raketka [301]
Pure magnesium's formula would just be Mg because all elements except for 7 nonmetals are just left alone when they are by themselves in a formula. The 7 diatomic elements( means they have to have two of them without another element attached to it aka. a subscript two after it when it's by itself) are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. An easy way to remember the diatomic seven is that when looking at a periodic table if you trace over them from nitrogen over to fluorine and down to iodine all of those elements are diatomic + hydrogen. 

And your unbalanced and balanced equations are correct.

(sorry I went on a tangent with the diatomic rules hopefully it will help you in the future though) 
8 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
Other questions:
  • HELP
    10·2 answers
  • The earth consist of how many chemical layers ?
    15·2 answers
  • When the following equation is balanced, the coefficient of H2O is Ca(s) + H2O + Ca(OH)2 + H2 a) 2
    15·1 answer
  • What of the following could be classified as matter
    13·1 answer
  • What is cerium’s common molecule name?
    15·1 answer
  • Using the following balanced equation:
    13·1 answer
  • PLS PLS PLS PLS HELP IM STRUGGLING
    15·1 answer
  • Which of these is an example of chemical change
    12·1 answer
  • Which of the liquids you tested (isopropyl alcohol, water, and glycerol) displayed the greatest surface tension (greatest interm
    5·1 answer
  • What does Fe(NO3)2 + Cu stands for?​
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!