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Andrew
11 days ago
7

A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft

drink? (Assume a density of 1.04 g/mL.)
Express your answer to three significant figures and include appropriate units.
Chemistry
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Lactose, a sugar in milk, is composed of one glucose molecule joined by a glycosidic linkage to one galactose molecule. how is l
lorasvet [2795]
The classification is disaccharide.

<span>Lactose qualifies as a disaccharide since it's made up of two monosaccharides, which are simple sugars (glucose and galactose). Disaccharides constitute a subgroup of carbohydrates, and besides lactose, other common examples include sucrose and maltose.</span>

8 0
2 months ago
A 0.821 gram sample of pure NH F was treated with 25.0 mL of 1.00 M NaOH
KiRa [2933]
A total of 0.0222 moles of NaOH are necessary to react with NH4F. \nBased on the reaction NH4F + NaOH --> NaF + NH3 + H2O, we start with: \nMass of NH4F = 0.821 g, NaOH concentration = 1 M, volume of NaOH = 25 mL. \nTo find moles: moles of NaOH = (CV)/1000. Thus, moles of NaOH = (1 * 25)/1000 = 0.025 moles of NaOH used. \nThe molar mass of NH4F is 37 g/mol, making moles of NH4F = 0.821 / 37 = 0.0222 moles. \nThis shows that NaOH is in excess, with 0.025 - 0.0222 = 0.0028 moles of NaOH remaining. Hence, 0.0222 moles of NaOH are needed to react with NH4F.
8 0
3 months ago
A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6
alisha [2963]
Specific heat refers to the quantity of heat a material can absorb or release to alter its temperature by one degree Celsius. To calculate specific heat, we apply the equation for the heat absorbed by the system. The heat taken in or released by a system can be expressed by multiplying the mass of the substance by its specific heat capacity and the change in temperature. The formula is:
Heat = mC(T2-T1)
By substituting the provided values, we can find C, the specific heat of the substance.
2510 J = 0.158 kg (1000 g / 1 kg)(C)(61.0 - 32.0 °C) C = 0.5478 J/g°C
8 0
3 months ago
Read 2 more answers
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