• ### What is the weight of 3.94 moles of carbon? 1. 2.37 × 1024 amu 2. 2.85 × 1025 amu 3. 2.85 × 10−21 amu 4. 47.3 amu 5. 2.37 × 10−22 amu?

3.94 x 12 x 6.022 x 10^23 amu = looks like answer #2
3.94 x 12 x 6.022 x 10^23 amu = looks like answer #2
1 answer · Chemistry · 1 hour ago
• ### Determine the mass of copper that has the same number of atoms as there are in 13.98 mg of potassium. Answer in units of mg.?

(13.98 mg)*63.55/39.10 = maybe 22 mg but use a calculator
(13.98 mg)*63.55/39.10 = maybe 22 mg but use a calculator
2 answers · Chemistry · 1 hour ago
• ### Chemistry help please? :)?

7 billion people working at 3600 peas per person per hour and 17 hrs/day = 4.3 x 10^14 peas per day. 6.022 x 10^23 peas/[(4.3 x 10^14 peas/day)(365 day/year)] = 3.8 x 10^6 years.
7 billion people working at 3600 peas per person per hour and 17 hrs/day = 4.3 x 10^14 peas per day. 6.022 x 10^23 peas/[(4.3 x 10^14 peas/day)(365 day/year)] = 3.8 x 10^6 years.
1 answer · Chemistry · 1 hour ago
• ### Which one has the greatest mass? 1. 3.59 mol O atoms 2. 0.0032 kg O2 molec 3. 3.5 g O atoms 4. 8.81 × 1023 O2 molec?

3.59 mol O atoms = about 64 grams. 8.81 x 10^23 O2 molecules = about 1.4 or 1.5 moles, so about 45 g. Answer #1 is the largest mass.
3.59 mol O atoms = about 64 grams. 8.81 x 10^23 O2 molecules = about 1.4 or 1.5 moles, so about 45 g. Answer #1 is the largest mass.
2 answers · Chemistry · 1 hour ago
• ### Find the mistake in the solution below. Explain how to fix the mistake and what the correct solution to the problem is.?

Three mistakes. The 3 in the denominator disappeared. The "n" terms got inverted. And m^3/m^3 is just 1. So the right answer is -4/(3n^5k^3).
Three mistakes. The 3 in the denominator disappeared. The "n" terms got inverted. And m^3/m^3 is just 1. So the right answer is -4/(3n^5k^3).
3 answers · Mathematics · 1 hour ago
• ### 1.00 kg of ice at -24.0°C is placed in contact with a 1.00 kg block of metal at 5.00°C. They come to equilibrium at -8.88°C.?

Equate "m c delta-T" for the two substances: (1.00 kg)(c-metal)(13.88C) = (1.00 kg)(2000 J/kgC)(15.12C) => c-metal = (2000 J/kgC)*(15.12 C)/(13.88 C) = about 2180 J/kgC. Actually, there are NO metals with heat capacities as high as that! Higher heat capacities tend to be associated with insulators like cork and peat. Oh well, not... show more
Equate "m c delta-T" for the two substances: (1.00 kg)(c-metal)(13.88C) = (1.00 kg)(2000 J/kgC)(15.12C) => c-metal = (2000 J/kgC)*(15.12 C)/(13.88 C) = about 2180 J/kgC. Actually, there are NO metals with heat capacities as high as that! Higher heat capacities tend to be associated with insulators like cork and peat. Oh well, not our problem.
3 answers · Physics · 5 hours ago

They are not saying whether the "t" is the time after midnight, or what ! I'll assume it's the time after midnight. Anyway, we're first looking for times when 3.5*cos[(pi/6)t] + 4.5 is more than 3.0. So we're looking for times when 3.5*cos[(pi/6)t] is algebraically greater than -1.5. If I set cos[(pi/6) t] = -1.5/3.5,... show more
They are not saying whether the "t" is the time after midnight, or what ! I'll assume it's the time after midnight. Anyway, we're first looking for times when 3.5*cos[(pi/6)t] + 4.5 is more than 3.0. So we're looking for times when 3.5*cos[(pi/6)t] is algebraically greater than -1.5. If I set cos[(pi/6) t] = -1.5/3.5, that implies (pi/6) t = +/- 2.01 radians, so t = 3.84 hours or 8.16 hours; and then the cycle repeats, so that diving becomes unsafe again at 3.84 hours after noon. So the safe diving times in daylight are from 8.16 hours after midnight to 3.84 hours after noon, which means from 8:10 a.m. to 3:50 p.m.
2 answers · Mathematics · 5 hours ago
• ### Is it true the Mars Expedition will only be able to take people with a body weight of less than 100lbs?

Lots of different manned missions to Mars are in the planning stages, none of them looking realistic. I don't know the one called "Mars Expedition," where did you learn about this?
Lots of different manned missions to Mars are in the planning stages, none of them looking realistic. I don't know the one called "Mars Expedition," where did you learn about this?
9 answers · Astronomy & Space · 6 hours ago
• ### Is it true? During the equinox, even though all latitudes have 12 hrs daytime and 12 hrs nighttime?

Carol is right to complain that the 12 hr day/12 hr night thing does not occur on EXACTLY the same date at all latitudes. The idea that it does, is based on assuming the earth's orbit to be circular and the earth to be spherical. So the "real" equinox in terms of a 12-hr day will vary by a few days from place to place. But as to the... show more
Carol is right to complain that the 12 hr day/12 hr night thing does not occur on EXACTLY the same date at all latitudes. The idea that it does, is based on assuming the earth's orbit to be circular and the earth to be spherical. So the "real" equinox in terms of a 12-hr day will vary by a few days from place to place. But as to the assertion about twilight, THAT would be true, because the apparent motion of the sun at the equator is nearly "straight down" after sunset, whereas at either of the poles, the sun would never be more than 23 degrees below the horizon (on the equinox day), and there would be several hours of "nautical twilight" and "nautical dusk" (periods when the sun is less than 12 degrees below the horizon). As a real example, consider Anchorage, Alaska, where the day length on March 18th will be 12:02, but the interval of light including both a.m. and p.m. twilights will be 15 hrs 12 min. In contrast to that, consider Honolulu, where the 12:00 day will occur on March 15th, but the length of light including both twilights will be only 13 hrs 36 min (or on March 18th, 13 hrs 40 min).
3 answers · Astronomy & Space · 8 hours ago
• ### Evaluate: 243^3/5 and 10x^3y^4(4x^-2y^-1)?

(a) 243^(3/5) = 3^3 = 27. (b) I guess you mean 10x^3y^4[4x^(-2)y^(-1)]. The product is 40xy^3.
(a) 243^(3/5) = 3^3 = 27. (b) I guess you mean 10x^3y^4[4x^(-2)y^(-1)]. The product is 40xy^3.
3 answers · Mathematics · 9 hours ago
• ### Physics Question?

(a) alpha = (vf - v0)/t = 625 rad/s^2. (b) 625 rad = 1875 cm = 18.75 m, so a = 18.75 m/s^2. (c) The centripetal acceleration is rw^2 = (0.03 m)(50 s^(-1)) ^2 = 75 m/s^2. Adding that to the tangential acceleration (which is perpendicular to it), the vector sum is about 77.3 m/s^2.
(a) alpha = (vf - v0)/t = 625 rad/s^2. (b) 625 rad = 1875 cm = 18.75 m, so a = 18.75 m/s^2. (c) The centripetal acceleration is rw^2 = (0.03 m)(50 s^(-1)) ^2 = 75 m/s^2. Adding that to the tangential acceleration (which is perpendicular to it), the vector sum is about 77.3 m/s^2.
1 answer · Physics · 9 hours ago
• ### Hlep solve?

They just wanted a capital V instead of a small v. We can't do part (b) without the given data.
They just wanted a capital V instead of a small v. We can't do part (b) without the given data.
2 answers · Mathematics · 9 hours ago
• ### I was trying to figure this out but i'm stuck and you guys help.?

Best answer: Line m passes through (4,1) and has a slope of -1/3, so y = (-1/3)x + 7/3. In "standard" form this becomes x + 3y = 7.
Best answer: Line m passes through (4,1) and has a slope of -1/3, so y = (-1/3)x + 7/3. In "standard" form this becomes x + 3y = 7.
2 answers · Mathematics · 9 hours ago
• ### Help in Volumes of Solid of Revolution?

I'll start by solving the two oblique lines simultaneously: y = (3x + 22)/5 = (20 - 5x)/3 => 9x + 66 = 100 - 25x => x = 1 and y = 5. The area is a triangle whose vertices are (6,8), (6,-50/3), and (1,5). The method of cylindrical shells will be convenient. Height of a shell = [(3x+22)/5 - (20-5x)/3]. Radius of a shell = x. Volume of a... show more
I'll start by solving the two oblique lines simultaneously: y = (3x + 22)/5 = (20 - 5x)/3 => 9x + 66 = 100 - 25x => x = 1 and y = 5. The area is a triangle whose vertices are (6,8), (6,-50/3), and (1,5). The method of cylindrical shells will be convenient. Height of a shell = [(3x+22)/5 - (20-5x)/3]. Radius of a shell = x. Volume of a shell = 2*pi*x*[(3x+22)/5 - (20-5x)/3] dx. = 2*pi[(34/15)x^2 - (34/15)x] dx So after integration you have (68/15)*pi*(x^3/3 - x^2/2), and the limits on the integral are x = 1 to 6. Evaluate: (68/15)*pi*(72 - 18 - 1/3 + 1/2) = 771.4 cubic units. So evidently I've made a mistake. See if you can find the mistake!
2 answers · Mathematics · 9 hours ago
• ### PHYSICS HOMEWORK HELP?

Without the diagram, we are only guessing. So here's my guess. The release point is 6.8 m above point A. So (1/2)mv^2 = mgh => v = sqrt(2gh) = sqrt(2*9.8 m/s^2*6.8 m) = 11.5 m/s.
Without the diagram, we are only guessing. So here's my guess. The release point is 6.8 m above point A. So (1/2)mv^2 = mgh => v = sqrt(2gh) = sqrt(2*9.8 m/s^2*6.8 m) = 11.5 m/s.
1 answer · Physics · 10 hours ago
• ### Show that vectors v=(2-1, 2t+1) and u=(t^2-1, 2t^2+3t+1) are parallel for all values of t≠1?

u = (t^2 - 1, 2t^2 + 3t + 1) = scalar (t+1) times vector (t-1, 2t+1). So vector u is a scalar multiple of vector v, so it has the same direction (i.e., is parallel). The reason why t = -1 should be excluded is that multiplying by zero creates anomalies. (In particular, "u" would then be the zero vector.) I can't see any reason to... show more
u = (t^2 - 1, 2t^2 + 3t + 1) = scalar (t+1) times vector (t-1, 2t+1). So vector u is a scalar multiple of vector v, so it has the same direction (i.e., is parallel). The reason why t = -1 should be excluded is that multiplying by zero creates anomalies. (In particular, "u" would then be the zero vector.) I can't see any reason to exclude t = +1.
1 answer · Mathematics · 10 hours ago
• ### What exactly does the y, m, x, and b stand for in y = mx + b?

"y" is the vertical distance between the x-axis and some point on the line. WHICH point on the line? -- the point whose "x" is the horizontal distance between the y-axis and the point of interest. "b" is a particular "y" value corresponding to x = 0.
"y" is the vertical distance between the x-axis and some point on the line. WHICH point on the line? -- the point whose "x" is the horizontal distance between the y-axis and the point of interest. "b" is a particular "y" value corresponding to x = 0.
6 answers · Mathematics · 10 hours ago
• ### Round off 75.135 to the nearest km?

If that's 75.135 km, then the nearest km is 75 km. If it was 75.135 m, then the nearest km is 0 km.
If that's 75.135 km, then the nearest km is 75 km. If it was 75.135 m, then the nearest km is 0 km.
5 answers · Mathematics · 10 hours ago
• ### Heat reactions? Chemistry?

The 2219.2 kJ would result from burning 1 mole of propane. The molar mass of propane is only 3*12 + 8*1.008 = 44.06 grams. So 286.4 grams of propane is 6.50 moles, and the energy released would be (6.50 mol)(2219.2 kJ/mol) = 14.4 MJ.
The 2219.2 kJ would result from burning 1 mole of propane. The molar mass of propane is only 3*12 + 8*1.008 = 44.06 grams. So 286.4 grams of propane is 6.50 moles, and the energy released would be (6.50 mol)(2219.2 kJ/mol) = 14.4 MJ.
2 answers · Chemistry · 11 hours ago
• ### Rational Zeros, Real Zeroes, and Imaginary Zeroes?

1. Obviously x = 0 is one of the zeros of this polynomial. So work with 12x^3 + x^2 - 4x - 2. Now the factors must begin with 1,2,3,4,6, or 12; and must end with 1 or 2. So the possible zeros are +/-2, +/-1, +/-1/2, +/-1/3, +/-1/4, +/-1/6, +/-1/12, +/-2/3. 2. Because the coefficient of the highest power is the largest coefficient, the whole... show more
1. Obviously x = 0 is one of the zeros of this polynomial. So work with 12x^3 + x^2 - 4x - 2. Now the factors must begin with 1,2,3,4,6, or 12; and must end with 1 or 2. So the possible zeros are +/-2, +/-1, +/-1/2, +/-1/3, +/-1/4, +/-1/6, +/-1/12, +/-2/3. 2. Because the coefficient of the highest power is the largest coefficient, the whole numbers are unlikely solutions, so I'll start with x = +/-1/2. In this case, the absolute values of the terms would be 3/2, 1/4, 2, 2, and no combination of signs will make them balance. Next I'll try x = +/- 1/4. The absolute values of the terms would be 3/8, 1/16, 1, 2, and no combination of signs will make them balance. Next I'll try x = +/- 1/3. The absolute values of the terms would be 4/9, 1/9, 4/3, 2, and these MIGHT balance...if x = -1/3 you'd have -4/9 + 1/9 + 4/3 - 2...nope, doesn't work. If x = +1/3 you'd have 4/9 + 1/9 - 4/3 - 2, also doesn't work. Now I'll try x = +/- 2/3. The absolute values of the terms would be 32/9, 1/9, 8/9, 2 Still won't balance with any sign combination. At this point I got lazy and went to a graphing calculator and found that the only real zero is around 0.71262, and cannot match any of the possible rational zeros. So there are no rational zeros (except for x=0). 3. If you divide 12x^3 + x^2 - 4x - 2 by the quantity (x - 0.71262), you'll get a quadratic whose coefficients are not exactly right, but that quadratic can then be solved to find approximate values for the imaginary zeros.
1 answer · Mathematics · 11 hours ago