Answer:
gravity increase when distance decrease,mass increase
gravity decrease vwhen distance decrease,mass decrease
Answer:
There are formed 98.05 g of MnCl₂
Explanation:
The reaction is this one:
MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂
First of all, determinate moles. Divide mass /molar mass
150 g / 36.45 g/m = 4.11 moles of HCl
Ratio between HCl and MnCl₂ is 4:1
4 moles of HCl produce 1 mol of Chloride
4.11 moles of HCl 'll produce (4.11 . 1)/ 4 =1.03 moles of chloride
Molar mass . Moles = Mass
Molar Mass MnCl₂ = 95.2 g/m
95.2 g/m . 1.03 moles = 98.05 grams
Yes , CH3OH - this is methanol
Answer:
1.22 x 10²⁵ molecules CO₂
Explanation:
To find the amount of molecules, you need to multiply the number of moles by Avogadro's Number. Avogadro's Number is a ratio which represents the amount of molecules per every 1 mole. It is important to arrange this ratio in a way that allows for the cancellation of units (since you are going from moles to molecules, moles should be in the denominator). The final answer should have 3 sig figs like the given value.
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
20.2 moles CO₂ 6.022 x 10²³ molecules
--------------------------- x -------------------------------------- = 1.22 x 10²⁵ molecules
1 mole
Answer : The value of 
is 28.97 kJ/mol
Explanation :
To calculate of the reaction, we use clausius claypron 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,
= vapor pressure at temperature 
= 462.7 mmHg
= vapor pressure at temperature 
= 140.5 mmHg
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![-21.0^oC=[-21.0+273]K=252K](https://tex.z-dn.net/?f=-21.0%5EoC%3D%5B-21.0%2B273%5DK%3D252K)
= final temperature = ![45^oC=[-41.0+273]K=232K](https://tex.z-dn.net/?f=45%5EoC%3D%5B-41.0%2B273%5DK%3D232K)
Putting values in above equation, we get:
![\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B140.5mmHg%7D%7B462.7mmHg%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B252%7D-%5Cfrac%7B1%7D%7B232%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D28966.6J%2Fmol%3D28.97kJ%2Fmol)
Therefore, the value of is 28.97 kJ/mol
