Answer:
See below
Step-by-step explanation:
An exothermic reaction tends to occur spontaneously because the products are more stable than the reactants.
Nature tries to get to the lowest energy state.
Answer options from an alternative source
- fructose
- lactose
- starch
- glucose
- cellulose
Answer:
- fructose -monosaccharide
- lactose - disaccharide
- starch - polysaccharide
- glucose - monosaccharide
- cellulose - polysaccharide
Explanation:
Monosaccharides are carbohydrates that are the simplest form of a sugar. They cannot be further broken down into smaller carbohydrates, and represent the basic building block for carbohydrates. Monosaccharides can form disaccharides, which are the sugar formed when two monosaccharides join together, or polysaccharides, which are chains of monosaccharides.
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BHBr%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BH_2%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
Given: ![-\frac{d[HBr]}{dt}]=0.301](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%5D%3D0.301)
Putting in the values we get:
![\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B0.301%7D%7B2%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
![+\frac{1d[Br_2]}{dt}=0.151](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D%3D0.151)
Thus the rate of appearance of 
is 0.151
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Use M x V = M' x V'
0.300 x V = 0.100 x 250
V = .......... ml
