which of the following compounds would be considered an electrolyte? a. c6h12o6 b. naoh c. co2 d. agcl

Marvin records scientific data about Lake Superior. For which property has Marvin forgotten to use an SI unit of measurement?

Illusion [34]

Answer: low temperature

Explanation:-

S.I or M.K.S is a system for defining physical units as meter, kilogram, second, ampere, kelvin or celcius, candela, and mole together with a set of prefixes to indicate multiplication or division by a power of ten for measuring length, mass, time , current, temperature and amount of substance respectively.

Given :

lake length = 563 kilometers = $563\times 10^3m$

High temperature = $76.5^0F$

Low temperature = $-6.3^0C$

annual precipitation =762 mm= $762\times 10^{-3}m$

Thus low temperature in units of Fahrenheit is not an S.I unit of measurement.

8 0

4 years ago

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Starch and cellulose are both produced by plants, yet one is easily digested by animals and the other is not. Discuss the differ

Nadusha1986 [10]

<span>Starch and cellulose have the same substance but different structures. They are both polysaccharides. The basic unit of a polysaccharide is the glucose. Glucose, which contains carbon, hydrogen, and oxygen, have two forms. The alpha-glucose with an alcohol group attached to carbon 1 is down and the beta-glucose with the alcohol group attached to carbon 1 is up. Starch is the alpha-glucose while cellulose is the beta-glucose. Starches are linked into a straight chain whereas the cellulose are connected like a pile of stack paper. When the human body eats starch, it can digest the starch but not the cellulose because it has no enzyme that can break it down. </span>

4 0

3 years ago

ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y

aleksandr82 [10.1K]

Answer:

The expression to calculate the mass of the reactant is $m = \frac{1.080kJ}{31.2kJ/g}$

Explanation:

<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

$1,080J.\frac{1kJ}{10^{3}J } =1.080kJ$