The answer is: mass is 40.17 kilograms.
d = 0.758 g/mL; density of fuel.
V = 14.0 gal; volume.
A gallon is a unit of volume in both the US customary and imperial systems of measurement. The US gallon is defined as 231 cubic inches (3.785 liters).
1 gal = 3785.41 mL.
V = 14 gal · 3785.41 mL:
V = 52995.74 mL.
m = 52995.74 mL · 0.758 g/mL.
m = 40170.77 g; mass of fuel.
m = 40170.77 g ÷ 1000 g/kg.
m = 40.17 kg.
Answer:
I think its B im not sure
but i hope this helps
<span>Group 1 can be characterized as atoms that have 1 electron in their valence shell. This is valuable when dealing with these questions, because the loss or gain of valence electrons is what defines ionic relationships. When group 1 elements form ionic bonds with other atoms, they are extremely likely to lose their valence electron, since the nucleus has a weaker pull on it than, say, a chlorine atom has on its 7 valence electrons. The weaker pull between the nucleus and the valence electron of group 1 elements means that the radius is high, since the electron is more free to move with less pull on it. This also means that the first ionization energy is low, since it takes relatively little energy for that electron to be pulled away to another atom.</span>
Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Answer:
Answer:
step 1:balance skeleton equation the chemical equation:
Zn +HNO3➔Zn(NO3)2+NO+H2O
step 2: identity undergoing oxidation or reduction
here
Zn➔Zn(NO3)2
Zn is oxidized from 0 to 2 in oxidation no.
HNO3➔NO
N is reduced from 5 to 2 in oxidation no
Step 3: calculate change in oxidation no.
change in oxidation no
in Zn=0-2=-2=2
in
N=5-2=3
Step 4: Balance it by doing crisscrossed multiplication
we get;
3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O
step 6:Balance other atoms except H & O
3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O
3Zn +2HNO3+6HNO3➔3Zn(NO3)2+2NO+H2O
finally: balance H
<em><u>3Zn +8HNO3➔3Zn(NO3)2+2NO+4H2O</u></em>
