Answer:
120mph
Explanation:
Google
divide the speed value by 1.467
or
176 times 60 second in a minute times 60 minutes in an hour
than divide by 5280 the amount of feet in a mile
The molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol, Option D is the correct answer.
<h3>What is enthalpy of Combustion ?</h3>
The energy released when a fuel is oxidized by an oxidizing agent is called enthalpy of Combustion.
It is given that
a 1.0 g sample of magnesium is burned to form MgO. in doing so, 25.5 kj of energy are released.
Molecular weight of Magnesium = 24.35g
24.35 g makes 1 mole of Mg
1g = 1/24.35
For 0.04 moles 25.5 kJ is released
for 1 mole 25.5 *1/.04
= 620 kj/mol
Therefore the molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol.
To know more about enthalpy of combustion
brainly.com/question/14754029
#SPJ1
C. KOH + HBr → KBr + H₂O
Explanation:
The two equations above illustrates the conservation of matter. The law of conservation of matters states that
"in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".
By this law, we understand that the amount of a particular matter we are starting with should be the one we end with.
KOH + HBr → KBr + H₂O
Conserving Product Reactants
K 1 1
O 1 1
H 2 2
We can see that the amount on both sides are the same.
Learn more:
Conservation of matter brainly.com/question/2190120
#learnwithBrainly
Answer: NO2, NO, and O2.
<span>Free radicals are toxic substances produced by the body. In normal circumstances,the body can neutralize but<span>
when the level of these substances is to much,they accumulate
and can generate diseases,
such as osteoporosis and cancer.</span></span>
Step 1: Write Imbalance Equation
CH₃CHO + O₂ → CO₂ + H₂O
Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.
CH₃CHO + O₂ → 2 CO₂ + H₂O
Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.
CH₃CHO + O₂ → 2 CO₂ + 2 H₂O
Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e
CH₃CHO + 5/2 O₂ → 2 CO₂ + 2 H₂O
Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.
2 CH₃CHO + 5 O₂ → 4 CO₂ + 4 H₂O
