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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
  • How do you use the important points to sketch the graph of # . . . - Socratic
    Please see the explanation below The equation is y=2x^2+2 Compare this equation to y=ax^2+bx+c The coefficient of x^2 is a=2 >0 So the curve is convex Now, calculate the intercepts with the y-axis x=0, =>, y=2 So a point is (0,2) Now, calculate the intercepts with the x-axis y=0, =>, 2x^2+2=0 =>, x^2+1=0 x^2=-1 There are no intercepts with the x-axis When x=-x, the curve does not change The
  • 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
  • 1545 subjects were given ginko 1524 placebo. In ginko . . . - Socratic
    1545 subjects were given ginko 1524 placebo In ginko group, 246 developed dementian in the placebo it was 277 with dementia Significance level=0 01 to test claim (ginko is effective in dementia) Can you test claim and construct confidence interval? Statistics
  • How many molecules of N_2O_3 are present in 28. 3 g of N_2O_3? - Socratic
    2 228 × 10^"23" molecules 1 mole contains 6 022140857 × 10^"23" (Avogadro's number) The molecular weight of N_2O_3 is: 76 01 g mol You have 28 3 g, so you have 28 3 76 01 = 0 37 mole Hence you have 0 37 × 6 022140857 × 10^"23" = 2 228 × 10^"23" You have 2 228 × 10^"23" molecules
  • 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




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