That depends on the size of the apple. 5 grams is quite a small difference. The best thing to do is get an apple and weigh it.
Carbohydrates in almondsHere are examples for dry roasted, oil roasted, blanched, and raw almonds. RAW0.2g of carbohydrate in one single raw almond6 carbohydrates in 1 oz or 28g of raw almonds (about 22-28 nuts)28 carbohydrates in 1 cup of raw whole almonds (5 oz or 142g)DRY ROASTED5 carbohydrates in 1 oz or 28g of dry roasted almonds, this will be 22-28 nuts27 carbohydrates in 1 cup of dry roasted whole almonds (4.9 oz or 139g)OIL ROASTED5 carbohydrates in 1 oz or 28g of oil roasted almonds, this will be 22-28 nuts28 carbohydrates in 1 cup of oil roasted whole almonds (about 5.5 oz or 157g)OTHER STYLES6 carbohydrates in 1 oz or 28g of blanched almonds8 carbohydrates in 1 oz or 28g of honey roasted unblanched almonds2 carbohydrates in 1 tablespoon of blanched slivered almonds29 carbohydrates in 1 cup of blanched whole almonds (5.2 oz or 145g)40 carbohydrates in 1 cup of honey roasted unblanched whole almonds (5.2 oz or 145g)For free carbohydrate charts to use as daily guides for either weight loss or weight maintenance, see the page links, further down this page, listed under Sources and Related Links..
Multiply the mass of the compound by the conversion factor based on the percent composition of the element in the compound
4NH3 + 6NO -> 5N2 + 6H2O 145 g N2 x (1 mol/28.0 g) x (6 mol NO/5 mol N2) x (30.0 g NO/1 mol NO) = 186.4285714 g NO rounds to 186 g NO due to significant figures
To determine the mass of glucose that must be metabolized to produce 145 grams of water, we can use the balanced equation for the combustion of glucose: (C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O). From the equation, 1 mole of glucose produces 6 moles of water. The molar mass of water (H₂O) is approximately 18 g/mol, so 145 g of water is about 8.06 moles (145 g / 18 g/mol). Therefore, to produce this amount of water, approximately 1.34 moles of glucose are needed (8.06 moles of water / 6). The molar mass of glucose (C₆H₁₂O₆) is about 180 g/mol, so the mass of glucose required is approximately 241.2 grams (1.34 moles × 180 g/mol).
According to NutritionData.com, in a 145g piece of gizzard, there are 212 calories.
Kinetic energy is 0.45 joules using the formula Ek = ½ x mv2
145g of Potassium Oxide ( K2O )contains a mass of 24.6279566g of Oxygen ( O )& a mass of 120.3705041g of Potassium ( K )24.6279566g O + 120.3705041g K2 = 144.9984607g K2O * Potassium Oxide (K2O)- 94.195 g/mol* Oxygen ( O )- 15.999 g/mol* Potassium ( K )- 39.098 g/mol ( K2 ) - 78.196145g x (1mol/94.196g) = 1.539343497mol K2O 145g K2O = 1.539343497 mol K2O Please note the concept: K2O has a Molar Ratio of 1 Oxygen:2 PotassiumSo... 1.539343497 mol K2O is representative of(1 x 1.539343497 mol) of Oxygen &(2 x 1.539343497 mol) of Potassium 1.539343497mol O x [ ( 15.999g O ) / ( 1mol O) ] = 24.6279566g O and 1.539343497mol K x [ ( 39.098 g K ) / ( 1mol K ) ] x [ ( 2mol K ) / ( 1mol K ) ] =120.3705041g K2 Please Note: the Molar Ratio ex: (2mol/1mol) etc. is the last step 24.6279566g O + 120.3705041g K2 = 144.9984607g K2O only a 0.001062% of error (some percent of error is always present no matter what)Percent Error is calculated by%Error= [ ( experimental value - accepted value ) ] x100%
145 grams NaCl (1 mole NaCl/58.44 grams)(1 mole Na +/1 mole NaCl)(6.022 X 1023/1 mole Na +) = 1.49 X 1024 ions of sodium =====================
The acute oral LD50 of alprazolamin rats is 2171 mg/kg.Assuming similar toxicity in humans (which is an assumption that has many faults) a 70 kg person would have a 50% of dying if they consumed about 145g that is over 500 0.25g tablets.
To calculate the force required to stop the ball, we need to convert the mass of the ball from grams to kilograms (145g = 0.145kg). Then, we can use the formula for force: force = mass x acceleration. Given that the ball is traveling at 100mph (44.7 m/s) and comes to a stop, the acceleration would be the time it takes to stop divided by the initial velocity. So, the force required to stop the ball would be calculated using these values.
I weighed 200 cubic centimetres of milk chocolate chips at a temperature of 22.5 C and the mass was 145g. That means that the density is 0.725g/cubic centimetre. However the density may vary if you are talking about dark or white chocolate chips. This is a measure of the density of Milk chocolate chips.
Just read that 5 oz (so 5/16 of 1 lbs) fits in 1 US cup (237 ml).
The contents of the jam jar is usually around 145 grams (i.e. not including the weight of the jar). This is usually because (although jam is a preserve, when it is made industrially it is not boiled to the extent that it would be when homemade, so is less "preserved" than it used to be), jam is not as long-life as it used to be. It is advised that it is kept in the fridge, so a 145g jar is usually the right size to fit in the fridge door.
Carbohydrates in almondsHere are examples for dry roasted, oil roasted, blanched, and raw almonds. RAW0.2g of carbohydrate in one single raw almond6 carbohydrates in 1 oz or 28g of raw almonds (about 22-28 nuts)28 carbohydrates in 1 cup of raw whole almonds (5 oz or 142g)DRY ROASTED5 carbohydrates in 1 oz or 28g of dry roasted almonds, this will be 22-28 nuts27 carbohydrates in 1 cup of dry roasted whole almonds (4.9 oz or 139g)OIL ROASTED5 carbohydrates in 1 oz or 28g of oil roasted almonds, this will be 22-28 nuts28 carbohydrates in 1 cup of oil roasted whole almonds (about 5.5 oz or 157g)OTHER STYLES6 carbohydrates in 1 oz or 28g of blanched almonds8 carbohydrates in 1 oz or 28g of honey roasted unblanched almonds2 carbohydrates in 1 tablespoon of blanched slivered almonds29 carbohydrates in 1 cup of blanched whole almonds (5.2 oz or 145g)40 carbohydrates in 1 cup of honey roasted unblanched whole almonds (5.2 oz or 145g)For free carbohydrate charts to use as daily guides for either weight loss or weight maintenance, see the page links, further down this page, listed under Sources and Related Links..
To find the equilibrium temperature, we can use the principle of conservation of energy. The total heat lost by the copper block as it cools down should be equal to the total heat gained by the aluminum calorimeter cup and the water as they warm up. This can be calculated using the formula: mcΔT = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. Once you solve this equation, you can find the equilibrium temperature at which heat transfer is balanced.
Multiply the mass of the compound by the conversion factor based on the percent composition of the element in the compound