You need to add the last substance to the products side(the right sode of the arrow). You have hydrogen and oxygen - water.
You get: BrO3 + N2H4 -> Br2 + N2 + H2O
# of Br: 1x1 = 1 # of Br: 2x1 = 2
O: 3x1 = 3 O: 1x1 = 1
N: 2x1 = 2 B N: 2x1 = 2
H: 4x1 = 4. H: 2x1 = 2
Br:
Multiply the reactant (left) side by 2 to balance.
O:
You've just multiplied the reactant oxygen by 2 so now the reactant side equals 6. Multiply the product (right) side by six as well.
H:
The product side is now equal to 12. Multiply the reactant side by 3 to balance.
N:
Now you have to balance N because the reactant side has been risen. So multiply the product side by three as well.
You end up with the complete and balanced equation:
2BrO3 + 3N2H4 -> Br2 + 3N2 + 6H2O
Answer:
14 J of heat
Explanation:
Recall that the enthalpy of reaction is defined as the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, if all reactants and products are in their standard states. Here we have the heat of formation of six moles of the substance and we are expected to use it to obtain the the heat of formation of one mole of the compound and we can do that by simple proportion as shown below;
If 6 moles of the compound produces 84J of heat
1 mole of the substance will produce 1×84/6 = 14 J of heat
Answer:
Thousands to Millions of Years
Explanation:
The rock cycle does not have just one proper direction, there are many pathways for rocks to travel through it. However, for most rocks to be changed from type to another will likely take anywhere from a few million to hundreds of millions of years.
Answer:
1.54 ml
2. 54 g
3. -43.7 °C
4. 97 g
5. 17.74 kJ
Explanation:
1. Volume of water that only came from the melted ice is;
Volume of water plus melted ice-volume of warm water
151 ml - 97 ml = 54 ml
2.Mass of water that came from the melted ice.
Mass=density of water * volume of water from melted ice
M=1 g/ml * 54 ml = 54 g
3. The change in temperature of warm water will be;
Lowest temperature of ice-water mixture-Temperature of warm water
ΔT=0.9-44.6 = -43.7
4. Mass of warm water will be;
Mass of warm water= density of water*volume of warm water
M=1g/ml * 97 = 97 g =0.097 kg
5.
Energy released by warm water as it cools
Q=mcΔT
Given , mass of warm water=m=97
c=4.184 j/g*°C = 4184 J/(kg.C)
Δ= - 43.7
Q=0.097*4184*43.7 =17735.56 J
17735.56/1000 =17.74 kJ
Answer:
it is regurgitated and an owl pellet