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- Question #b5f73 - Socratic
Then, we can just find the mass of 2 272 carbon atoms by multiplying 2 272 by the mass of 1 mole of carbon: 2 272 xx 12 01 = "27 29g" Since there was 1 significant figure in the question, our answer is "30g" Another way to solve this problem is by using the mass percent!
- Question #7908b - Socratic
65 0 "g N"_2 Any time you want to convert between moles and grams of a substance, you must know that substance's molar mass, which is the mass of one mole of that substance (One mole is equal to Avogadro's number (6 022xx10^23) of individual particles molecules of that substance ) To find the molar mass of a nitrogen molecule, we'll first realize its chemical formula, which is "N"_2 Most
- Question #004f2 - Socratic
0 56 1 molar solution ="Number of moles of solute" "Volume of solution" for Sugar C_ (12)H_ (22)O_ (11) molar mass=342gm according to equation and information provided 1= (3 42 342) x where x is the volume of solution in l : the volume of solution is 0 01 l which is 10 ml : 10 gm of water moles of solute=3 42 342=0 01 moles of solution =10 18=0 55 : the mole fraction of the solution is 0
- How many grams are there in 7. 8x1019 atoms of Beryllium (Be)?
Well, we know that 6 022xx10^23 beryllium atoms have a mass of 9 01*g How do we know this well 6 022xx10^23 specifies Avogadro's number, and the atomic masses are conveniently printed on the Periodic Table that is conveniently beside you
- Question #18488 - Socratic
The degree of dissociation sf (alpha=0 0158) sf (K_b=2 51xx10^ (-6)color (white) (x)"mol l") Triethyamine is a weak base and ionises: sf ( (CH_3)_3N+H
- Question #5d6ba - Socratic
x~~163 43 I am interpreting it as log_10 (0 01000^2 x^3)=-10 64 Remember that log (a b)=log (a)-log (b) Thus, log_10 (0 01000^2)-log_10 (x^3)=-10 64 Now, log (a^b)=blog (a) Then, 2log_10 (0 01000)-3log_10 (x)=-10 64 Since 10^-2=0 01, log_10 (0 01)=-2 Then, (2*-2)-3log_10 (x)=-10 64 Simplify and rearrange the terms to get 3log_10 (x)=-4+10 64=6 64 Divide both sides by 3: log_10 (x)=6 64
- Question #d407c - Socratic
The solutions are S= {160 6^@,305 7^@} The max and min are =5 and -5 rcos (theta+alpha)=r (costhetacosalpha+sinthetasinalpha) =rcosthetacosalpha+rsinthetasinalpha) Comparing this to the equation 3costheta-4sintheta We see that rcosalpha=3, and rsinalpha=-4 Therefore, (rcosalpha)^2+ (rsinalpha)^2=r^2 r^2=3^2+ (-4)^2=9+16=25 Therefore, r=sqrt (25)=5 cosalpha=3 5 and sinalpha=-4 5 alpha=-53 13
- Question #cbccf - Socratic
The concentration of "Pb"^"2+" at equilibrium is 9 3 × 10^"-5"color (white) (l) "mol L" > We can solve this by setting up an ICE table color (white) (mmmmmm)"Pb (s
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