Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j
Answer is: specific gravity of glucose is 1,02.
d(glucose) = 1,02 g/ml.
d(water) = 1,00 g/ml.
Specific gravity of glucose = density of glucose ÷ density of water.
Specific gravity of glucose = 1,02 g/ml ÷ 1,00 g/ml.
Specific gravity of glucose = 1,02.
Specific gravity<span> is the ratio of the </span>density<span> of a substance (in this case glucose) to the density of a reference substance (water).</span>
Answer:
what would u like me to do on this?
Explanation:
do you have a tablel of elements? if not then look at on online and search them all.
Answer:
Yes
Explanation:
A reaction normally takes place between metals and non metals. The metals acts as the electron donors and the non metals acts as the electron acceptors. This exchange of electrons form bonds such as ionic or covalent.
A good example of a reaction between a metal and non metal is Sodium metal and Chlorine(non metal). They form an ionic bond and the product is Sodium chloride.
