Shred red cabbage ~ (3/4 of a very small head)
Put the cabbage pieces in a small container ~ ( you can use a Pyrex-4-cup measure, a bowl or even a plastic zipper bag)
Cover the cabbage with very hot water. Let it sleep until the water has cooled. (somewhere between lukewarm and room-temperature)
The purple liquid you've made is your indicator.
Pour it into a container and compost the cabbage.
Now look for substances that may be acids or bases.
Liquids are good, like fruits.
You can also use solids around for baking are good too. (such as baking soda, salt, sugar, cream of tartar...)
Get containers for mixing (such as tea cups, because they are small, shallow and white inside)
Pour the indicator into the tea cups and add an acid or base.
Lemon juice, rice wine vinegar, and apple cider vinegar, turn the cabbage-water indicator into a pink.
Orange juice or fresh oranges (same thing) turn the cabbage-water indicator into an orangish-pinkish color.
Baking soda turns the cabbage-water indicator blue.
Milk (non-fat) turns the cabbage-water indicator turn opaque and milky, yet purple.
An egg white (which won't get into the solution immediately until after a lot of stirring) turns the cabbage-water indicator blue.
Hint:
Bases mostly turn the indicator towards blue-ish colors such as purple, light blue, dark blue, opaque blue...
Acids mostly turn the indicator towards pink-ish colours such as orange-ish pink, floral pink...
(You'll have to keep on testing the cabbage-water indicator in after a day or two to see if the indicator quality persists or degrades.
Answer:
223 g O₂
Explanation:
To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.
4 Al + 3 O₂ ----> 2 Al₂O₃
^ ^
Molar Mass (O₂): 32.0 g/mol
9.3 moles Al 3 moles O₂ 32.0 g
------------------- x --------------------- x -------------------- = 223 g O₂
4 moles Al 1 mole
Answer:
The given atom is of Ca.
Explanation:
Given data:
Speed of atom = 1% of speed of light
De-broglie wavelength = 3.31×10⁻³ pm (3.31×10⁻³ / 10¹² = 3.31×10⁻¹⁵ m)
What is element = ?
Solution:
Formula:
m = h/λv
m = mass of particle
h = planks constant
v = speed of particle
λ = wavelength
Now we will put the values in formula.
m = h/λv
m = 6.63×10⁻³⁴kg. m².s⁻¹/3.31×10⁻¹⁵ m ×( 1/100)×3×10⁸ m/s
m = 6.63×10⁻³⁴kg. m².s⁻¹/ 0.099×10⁻⁷m²/s
m = 66.97×10⁻²⁷ Kg/atom
or
6.69×10⁻²⁶ Kg/atom
Now here we will use the Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Now in given problem,
6.69×10⁻²⁶ Kg/atom × 6.022 × 10²³ atoms/ mol × 1000 g/ 1kg
40.3×10⁻³×10³g/mol
40.3 g/mol
So the given atom is of Ca.
Evaporation,condensation,precipitation,sublimation,transpirtation,runoff and infiltration
The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
The energy of the photon is given by;
E = hc/λ
E = energy of the photon
h = Plank's constant
c = speed of light
λ = wavelength of light
For the upper boundary range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 270 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9
E = 7.33 × 10^-19 J
For the lower range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ =300 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9
E = 6.6 × 10^-19 J
Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
Learn more: brainly.com/question/24857760
