Answer:
Alpha, Beta, and Gamma particles according to ability to penetrate matter.
Explanation:
- Alpha represent the largest particle which penetrates the least with a positive charge,
- Beta particles carry a negative charge, and
- Gamma rays are neutral.
Conduction-
putting an icepack on injury
grabbing a warm coffee mug
burning yourself by touching boiling water
the handle of a pot being to hot to touch
Convection-
heat from a fire to warm hands
warm water rising to the surface of of the ocean etc.
warm air rising off pavement
an oven that cooks by cycling warm air through the bottom and out the top
Radiation-
cooking popcorn using microwave
heat from the sun hitting a solar panel
( i havent done science in a while cuz im homeschooled but i think most of these are correct)
good luck on what your working on !!
B. horizontally Thats the answer
Answer:
Molarity = 2.3 M
Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution
1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles
2- The volume of solution is given = 1 liter
3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M
Hope this helps :)
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>