What is the mass of 15.00 mL of pure water at 20 C?
96826
By the original definition, the mass of one mL of water is one Gram (temperature dependent, but really, really close at 20 degrees C.
249474
The mass of 1 cm^3 of water is 1 gramme.  Therefore, 100 cm^3 of water will have a mass of 100 g. If you add 150 g of sugar (or salt or anything else that does not produce an explosion or gives off a gas) the total mass will be 250 g.\n\nIf you had asked what is the total volume, that would have been a completely different situation.
482741
You need a few different constants to figure this out.\n\nWater has a Specific Heat of 1 kcal/kg/°C\nThat means it needs 1 kcal per kilogram to heat the water 1 degree celsius\n\nIt has a density of 1000 kg/m3\n\nIt has a Latent heat of Fusion of 80 kcal/kg (melting) and 65 J/kg (freezing)\n\nSo if you know the volume you can work out the weight of the water. Once you have the weight (or you already know it), you can work out the rate of the drop (or increase) of water temperature.\n\nOnce you get to the point of freezing, or melting, simply add on the latent heat, and you can then get the total amount of energy required to change the water from one temperature to the next.\n\nSo for example;\n\n1) How much energy to heat 2 kg water from 18°C to 100°C is =>\n\n2kg * 1 * (100-18) = 164 kcal (The heat up curve is fairly straight)\n------------------------------->\n\n2) 2kg of water from 5°C to -10°C is as follows;\n\n(2 x 1 * (5 - -10)) + (65 * 2) = 160 kcal\n\nCurve drops down fairly constantly, then flats out until the water turns to ice, then continues to drop until you are at your temperature.\n\nI hope that helps a little.\n\nIf you want to figure out the times for all this to happen, you need to either know what heat source you are using (for example the wattage of the kettle element), and also you need to figure out the heat losses from the container (e.g. kettle or pipe, etc.)\n\nThese can all then be converted to an equivalent kcal number.
743809
U.S. customary units\n150 fl. oz. = 1.171875 gal.\n\n\nImperial unit (previously used by U.K. before replaced by metric\n150 fl. oz. = ~1.066666
492398
normally, liter is used to measure liquids and grammes for measuring weight. however, in the case of water and milk, 2100 ml would amount to 2.1 kg.
362368
Given:\nm <mass of substance> = 12.0g or 12/1000 kg\nc <specific heat capacity of water> = 4180J per kg per degree celsius\nt <change in temperature> = 20 degrees to 100 degrees i.e. 80 degree celsius\n\nFind:\nHeat req.\n\nSolution:\nHeat required or Q\nQ = m * c * t\n  = 12/1000 * 4180 * 80\n  = 12/10 * 418 * 8 (cancelling 0s of 4180&80)\n  = 12/5 * 209 * 8 (div. 418&10 by 2)\n  = 12/5 * 1672 (mult. 207&8)\n  = 20064/5 (mult. 12&1672)\n  = 4012.8\n\nAnswer = 4012.8 Joules or 4.0128 KJ
134374
dimensionally, volume is in units 1/(length)^3 and mass is in kilogrammes. So the mass of water is not strictly speaking equal to its volume. However, at standard temperature and pressure, 1 litre of water is equal to 1 kg of water. This is simply a definition (in the same way the temperature scale Celsius is 0C for freezing point and 100C for boiling point). So the reason why - is that it is defined as being equal; there is no particular physical reason for this.
127092
It would depend on the fluid that you are weighing.  Water for example is 1 g/mL at 20 degree Celcius. Other fluids with different densities will convert differently.
567447
The volume of the tank is 40 x 30 x 15 = 18000 cubic centimeters, which is the same as 18000 milliliters (mL).  12 Liters = 12000 milliliters, so it will take 18000/12000 = 1.5 minutes.  I am a high school math teacher.
788167
c. 1 mg/L\n\nbecause there is 1000000 mg in L ( water)
896861
Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
869561
C57H110O6(l)+109O2(g)--heat>\n57CO2(g)+55H2O(l)\n\n\nabout  15.05kg of  CO2
251863
At 20 degrees C, it's 0.6g/100mL. You can convert that to a Ksp.
1068631
If you mean ml to mg then the answer varies on the density of the liquid but ..\n1 ml of water = 1 mg of water
980842
See, first of all you need to convert your Celcius to Kelvin by the formula, 273+C=K. Therefore 20C becomes 293 K and -80C becomes 193. Now use this combined gas equation:\nP1V1/T1=P2V2/T2 where P1 is the initial pressure, V1 is the initial volume and T1 is the initial temperature. \nP1=335mm Hg\nV1=300ml\nT1=293K\n\nwhereas P2,V2 & T2 stand for final pressure, final volume (the one you have to find in this question), and final temperature.\nP2=760mm Hg, V2=x, T2=193K\n\nso substitute these values now,\n \n335*300/293 = 760*x/193\n\nx=335*300*193/760*293\nx=87.1ml
165248
(.002mol /20ml) (1000ml / 1L ) = .10 M NaOH\n\n\n pOH =  -log 0.10 =  1\n\n pH = 14.00 - pOH   = 14 - 1  = 13 = D
554942
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J