Answer:
197 + (35.5×3) = 303.5
Explanation:
relative formula mass is the sum of the relative atomic masses of the atoms in the formula ( AuCl3 )
Answer:
Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.
I choose question 1, so molarity is the concentration of a soulution expressed as the number of moles of solute by the litress of soulution. to get molarity you divide the moles of soulute by the litress of solution. soo 1 calculate the number of moles of solute present. 2 Calculate the number of litress solution present. 3. divide the number of moles of solute by the number of litress of solution
soo 1 mol of NaOH has a mass of 40.00 g, so moles of NaOH= 26.7. 1 mole divided 40.00 = 0.375. litress of solution = 650 g. im not sure why its a g i usally do it Ls so i guess its the way your teacher wants you to do it so do you know how to do that. so molarity = moles of solute and litress solution. sorry this probably didnt help i just wanted to add something that might help. im still working on this stuff myself hope this helps.
Answer:
B. Gravity held the pieces of forming planets together.
c. Gravity pulled most of the matter into the center of the solar system
D. Gravity caused the planets and Sun to have spherical shapes.
Explanation:
When a collection of grains pulled together by their gravitational forces would keep in by the gravity of a star, it would eventually became bigger to the point a planet was formed.
The sun's strong gravitational force pulled most of the matter around it to the center of the solar system.
The spherical shape of planets is a result of their gravity pulling equally from all sides, shaping it into a sphere.
Mass of CO₂ evolved : 0.108 g
<h3>Further explanation</h3>
Given
1.205g sample, 36% MgCO3 and 44% K2CO3
Required
mass of CO2
Solution
0.36 x 1.205 g=0.4338 g
mass C in MgCO₃(MW MgCO₃=84 g/mol, Ar C = 12/gmol)
= (12/84) x 0.4338
= 0.062 g
0.44 x 1.205 g = 0.5302 g
Mass C in K₂CO₃(MW=138 g/mol) :
= (12/138) x 0.5302
= 0.046 g
Total mass Of CO₂ :
= 0.062 + 0.046
= 0.108 g
