Answer:
the point at there the water is changing is titled "getting dressed"
The appropriate response is the fourth one. The announcement is valid about this condition beneath is in spite of the fact that it is unequal, it can be adjusted by specifically utilizing observer particles.
I hope the answer will help you.
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
The molar mass of methylammonium bromide is 111u.
<h3>What is molar mass?</h3>
The molar mass is defined as the mass per unit amount of substance of a given chemical entity.
Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.
Add it all together and put units of grams/mole after the number.
Atomic weight of H is 1u
Atomic weight of N is 14u
Atomic weight of C is 12u
Atomic weight of Br is 79u
Calculating molar mass of
=2(1 x3+ 14+12+ 1 x 3 +79) = 111u
Hence, the molar mass of methylammonium bromide is 111u.
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Answer:
H. 2 blue, 3 yellow, and 12 green
Explanation:
Aluminium atoms (Al) = Blue Beads
Oxygen Atoms (O) = Green Beads
Sulfur (S) = Yellow beads
From the compound Al2(SO4)3, the number of atoms present are;
Al = 2
S = 3
O = 12
This means the model would contain;
2 Blue beads
12 Green beads
3 Yellow beads
The correct option is; H. 2 blue, 3 yellow, and 12 green