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
muminat
2 years ago
9

How do robots help scientists to perform experiments?

Chemistry
2 answers:
Andreas93 [3]2 years ago
8 0

Answer:

A. Robots go places humans are unable to go.

Arte-miy333 [17]2 years ago
3 0
The answer is number A
You might be interested in
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
When 20 grams of KCIO3, is dissolved in 100 grams of water, the solution can be
tia_tia [17]

Answer:

b. unsaturated .

Explanation:

Hello there!

In this case, according to the given information, it turns out necessary for us to bear to mind the definition of each type of solution:

- Supersaturated solution: comprises a large amount of solute at a temperature at which it will be able to crystalize upon standing.

- Unsaturated solution: is a solution in which a solvent is able to dissolve any more solute at a given temperature.

- Saturated solution can be defined as a solution in which a solvent is not capable of dissolving any more solute at a given temperature.

In such a way, since 20 grams of the solute are less than the solubility, we infer this is b. unsaturated, as 33.3 grams of solute can be further added to the 100 grams of water.

Regards!

6 0
3 years ago
What effect does heating a solid to a hightemperature have have on its crystal lattice ??
kotegsom [21]

Arrangement of atoms or group of atoms in a three dimensional ordered pattern in a crystal is said to be a crystalline lattice. They are arranged in a specific pattern with high symmetry. The heating of the crystal at high temperature will result in the increase of vibrational kinetic energy of the atoms in the crystal and this will result in the breaking of lattice apart and due to the breaking of lattice apart the ions will flow freely.

Thus, the heating of a solid at high temperature will lead to the lattice breaks apart and ions will flow freely.

6 0
3 years ago
Read 2 more answers
A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,
swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

                      B                                   N                         H

mass          40.28 g                         52.20 g                 7.53 g

number of moles  

                 40.28 g / 11 g/mol      52.20 g / 14 g/mol    7.53 g / 1 g/mol

                = 3.662 mol                  = 3.729 mol             = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

                3.662 / 3.662              3.729 / 3.662              7.53 / 3.662

               = 1.000                           = 1.018                        = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂          

that can be written as BNH₂    

3 0
3 years ago
If the mass of a particular atom is exactly 5 times the mass of an atom of carbon-12, what is its mass
otez555 [7]
The mass is 60 grams.
8 0
4 years ago
Other questions:
  • The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie
    14·1 answer
  • What is the formula of the ionic compound formed by the elements lithium and oxygen?
    14·2 answers
  • Hormones are chemicals secreted by the
    5·1 answer
  • Which percentage is in the appropriate range for a snack to be considered a “good source” of nutrients?
    11·2 answers
  • Is cloud seeding a physical or chemical process?
    8·1 answer
  • Refer to Animation: Kinetic and Potential Energy. Which of the factors accounts for the higher potential energy of C-C and C-H b
    7·1 answer
  • The pencil is material or mixture ?please help ​
    13·1 answer
  • When a surface is experiencing friction with another surface, how are the particles
    11·1 answer
  • A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming
    14·1 answer
  • Calculation: If a compound is 14.40% hydrogen by mass and 85.60% carbon by mass, what is the empirical formula?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!