The atomic number in an element is usually how many protons the element has. For example, Hydrogen has a 1 on top of the H (on the periodic table), therefore, Hydrogen has 1 proton. Oxygen has an 8 on top of the O (on the periodic table) so therefore, Oxygen has 8 protons.
Answer:
A. tungsten
Explanation:
Tungsten is a material which high melting point ie. does not melt easily incase of high temperature
Answer:
N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e
Explanation:
Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.
The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.
The answer is 1.75 x 10¹⁰ m.
Scientific notation<span> refers to a mathematical expression that is used to represent a decimal number between 1 and 10 multiplied by ten, or we can say that to write a large numbers using less digits.
Diameter of chlorine atom = </span><span>175 picometers (pm)
</span>175 pm = <span>0.000000000175 m
in scientific notation we will write </span>0.000000000175 m as <span>1.75 x 10</span>¹⁰ m
<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm
<u>Explanation:</u>
To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= pressure of the liquid = ?
= Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = 341.88 K
= final temperature = 305.03 K
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28900J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B341.88%7D-%5Cfrac%7B1%7D%7B305.03%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-1.228atm%5C%5C%5C%5CP_2%3De%5E%7B-1.228%7D%3D0.293atm)
Hence, the vapor pressure of the liquid is 0.293 atm
