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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 #b5e69 - Socratic
Explanation: #M_ (CO_2) = "220 mg"# #MW_ (CO_2) = "44 01 g mol"# #"O"_2 " molecules = ?"#
- Question #6f7d1 - Socratic
675N Here,the kinetic energy of the hammer head got totally converted to the work done in moving the nail through a distance of 10 mm or 0 01 m So, kinetic energy of the hammer during the strike was 1 2 m v^2=1 2×1 5×3^2=6 75J Now,if an average force of F had acted on the nail to move it by 0 01 m,then work done by the force will be (F×0 01)J(using W=F*s) So,we can equate both 6 75=F×0 01
- 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
- 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
- Can anyone please answer this chemistry difficulty? - Socratic
E^@ = 0 Q = ( ["Zn"^"2+"]_text (prod)) ( ["Zn"^"2+"]_text (react)) = (0 01 color (red) (cancel (color (black) ("mol L")))) (1 color (red) (cancel (color (black) ("mol L")))) = 0 01 E = E^@ - (RT) (zF)lnQ = - ("8 314 V"·color (red) (cancel (color (black) ("C·K"^"-1""mol"^"-1"))) × 298 15 color (red) (cancel (color (black) ("K")))) ("2 ×
- How do you solve X * 0. 01 = 637? | Socratic
Multiply each side by color (red) (100) X * color (red) (100) * 0 01 = color (red) (100) * 637 X * 1 = 63700 X = 63700
- Question #fe281 - Socratic
4 ⋅ 12 01 = 48 04 g We now know that one mole of carbon contains 48 04 grams of carbon Next, we determine how many grams of carbon are in one mole of CO 1 ⋅ 12 01 = 12 01 g Now, we know that one mole of CO contains 12 01 grams of carbon With this information, we can determine how many moles of CO can be formed with the given mass of C
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