Answer:
410.196 J/[kg*°C].
Explanation:
1) the equation of the energy is: E=c*m*(t₂-t₁), where E - energy (523 J), c - unknown specific heat of copper, m - mass of this copper [kg], t₂ - the final temperature, t₁ - initial temerature;
2) the specific heat of copper is:
![c=\frac{E}{m*(t_2-t_1)}; \ => \ c=\frac{523}{0.085*(45-30)}=\frac{523}{1.275}=410.196[\frac{J}{kg*C}].](https://tex.z-dn.net/?f=c%3D%5Cfrac%7BE%7D%7Bm%2A%28t_2-t_1%29%7D%3B%20%5C%20%3D%3E%20%5C%20c%3D%5Cfrac%7B523%7D%7B0.085%2A%2845-30%29%7D%3D%5Cfrac%7B523%7D%7B1.275%7D%3D410.196%5B%5Cfrac%7BJ%7D%7Bkg%2AC%7D%5D.)
Answer:
As you move across a period, the atomic mass increases because the atomic number also increases. ... The atomic mass for any given atom mainly comes from the mass of the protons and neutrons.
Explanation:
Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
