Answer:
25.157 cm³
Explanation:
Data Given:
Mass of Sugar (m) = 40g
Density of sugar given in literature = 1.59 g/cm³
Volume of Sugar = ?
The formula will be used is
d = m/v ........................................... (1)
where
D is density
m is the mass
v is the volume
So
Rearrange the Equation (1)
d x v = m
v = m/ d ................................................ (2)
put the given values in Equation (2)
v = 40g / 1.59 g/cm³
v = 25.157 cm³
volume of 40 g of sugar = 25.157 cm³
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Answer:
D) forms hydronium ions in water
Explanation:
Texture and conductivity are physical properties which makes answer A and B wrong. Answer C does not refer to a chemical reaction.
Answer: Every enzyme has a specific name that can give us insight into the specific reaction that that enzyme can catalyze. We divide them into six different categories.
1) Oxidoreductase - includes two different types of reactions by transferring electrons from either molecule A to B or vice versa. It is involved in oxidizing electrons away from a molecule.
2) Hydrolase - uses water to divide a molecule into two other molecules.
3) Transferase - you move some functional group X from molecule B to molecule A
4) Ligase - catalyzes reactions between two molecules, A and B, that are combining to form a complex between the two. (example: DNA replication)
5) Lyase - divides a molecule into two other molecules without using water and without reducing or oxidation
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)
