Answer:
it will all fall unless these is air
Explanation:
Answer:
If you do it carefully enough, a small amount of water will pour out of the glass — that's supposed to happen. But try not to let any air bubbles get into the glass. Finally, slowly remove the hand holding the cardboard in place.
What is the frequency of a wave that has a wavelength of 0.50 m and a speed of 380 m/s?
Answer: f = 760 Hz
Explanation: speed = frequency · wavelength v = fλ.
frequency f = v/ λ = 380 m/s / 0.50 m = 760 Hz
Use the given Nernst equation and reaction to solve this problem. What is the potential of this cell with the given conditions?
2Li (aq) + F2(g) 2Li+(aq) + 2F- (aq)
E° = +5.92 volts
T = 200°C
[Li+] = 10.0 molar
[F-] = 10.0 molar
Answer:
The 2nd one is the one
Explanation:
and it isn't writen out all the way
Meera added blue copper sulphate crystals to some water in a beaker.
The copper sulphate dissolved in the water.
1 give one way meera could see that the copper sulphate had dissolved in the
Answer:
The solid crystals disappeared
Explanation:
When a soluble solid solute is added to water, the solid solute disappears after a little while. The disappearance of this solute indicates that the solute has been dissolved in water.
In this case, blue copper sulphate crystals are added to water, the blue crystals disappear leaving only a blue solution. The disappearance of these blue copper sulphate crystals indicates that the substance has dissolved in water.
How would you find the density of a can of soda pop?
A. Find the mass of the can of soda pop and then multiply by the number of cubic centimeters in the can
B. Find the mass of the can of soda pop and then divide by the number of cubic centimeters in the can
C. Convert a gallon into cubic centimeters and then divide by the mass of the can of soda pop
D. Convert a gallon into cubic centimeters and then subtract the mass of the can of soda pop
Answer:
it's A.
Explanation:
have uh good day ma :)))))))
what does the period number on the periodic table tell us?
Answer:
The period number on the Periodic table tells you the total number of orbits that the atom will have. In other words, the period number indicates the number of energy levels (or energy orbit) of an atom. For example, 1st period indicates that these elements possess 1 energy shell (or energy orbit).
Explanation:
why do we need to rinse the mouth before collecting the saliva
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(11) Deduce the number of dyes in food colouring H.
(iii) Suggest why food colouring F does not move during the experiment.
(iv) Explain which two food colourings contain the dye that is likely to be the most soluble the solvent.
(b) Determine which food colouring contains a dye with R, value closest to 0.67
Show your working.
Answer:
(ii) 1 dye
(iii) Food coloring F is insoluble in the solvent
(iv) 'E' and 'H'
(b) Food colouring G
Explanation:
Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper
The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper
(ii) The number of components into which the food colouring 'H' separates into = 1
Therefore, the number of dyes in food colouring 'H' = 1 dye
(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase
(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;
Food coloring 'E' and 'H'
(b) Using a similar question solution found on 'tutor my self' website, we have;
The [tex]R_f[/tex] values are given as follows;
[tex]R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}[/tex]
The distance moved by the solvent = 5 units
The distance moved by dyes in food colouring 'E' and 'H' = 4 units
The distance moved by dye in food colouring 'G' = 3.3 units
The distance moved by the second dye in food colouring 'E' = 2.7 units
By inspection, we get;
[tex]R_f[/tex] dye in food colouring 'G' = 3.3/5 = 0.66,
Therefore, the dye with [tex]R_f[/tex] value closest to 0.67 is the dye in food colouring 'G'.
5). At what temperature (K) will 0.854 moles of neon gas occupy 12.3 L at 1.95
atmospheres?
Answer:
338.38 K
Explanation:
Applying,
PV = nRT............... Equation 1
Where P = pressure, V = Volume, R = Temperature, n = number of moles, T = temperature.
Make T the subject of the equation,
T = PV/nR............. Equation 2
From the question,
Given: P = 1.95 atm, V = 12.3 L, n = 0.854 moles
Constant: R = 0.083 L.atm/K.mol
Substitute these values into equation 2
T = (1.95×12.3)/(0.854×0.083)
T = 338.38 K
How many moles of aluminum are in 54 grams
What are the two limitations of earth plates
Answer:
The tectonic style and viability of modern plate tectonics in the early Earth is still debated. Field observations and theoretical arguments both in favor and against the uniformitarian view of plate tectonics back until the Archean continue to accumulate. Here, we present the first numerical modeling results that address for a hotter Earth the viability of subduction, one of the main requirements for plate tectonics. A hotter mantle has mainly two effects: 1) viscosity is lower, and 2) more melt is produced, which in a plate tectonic setting will lead to a thicker oceanic crust and harzburgite layer. Although compositional buoyancy resulting from these thick crust and harzburgite might be a serious limitation for subduction initiation, our modeling results show that eclogitization significantly relaxes this limitation for a developed, ongoing subduction process. Furthermore, the lower viscosity leads to more frequent slab breakoff, and sometimes to crustal separation from the mantle lithosphere. Unlike earlier propositions, not compositional buoyancy considerations, but this lithospheric weakness could be the principle limitation to the viability of plate tectonics in a hotter Earth. These results suggest a new explanation for the absence of ultrahigh-pressure metamorphism (UHPM) and blueschists in most of the Precambrian: early slabs were not too buoyant, but too weak to provide a mechanism for UHPM and exhumation.
Explanation: