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My name is Ann [436]
3 years ago
8

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

Chemistry
1 answer:
never [62]3 years ago
4 0
C4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g
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Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g) A student combined two colorless aqueous solutions. One of the solutions contained N
Scrat [10]

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Explanation:

As it is given that both Na_{2}CO_{3} and HCl chemically combine together leading to the formation of NaCl, water and carbon dioxide gas. As the gas is forming and its formation will also form bubbles into the solution.

This formation of bubbles actually indicate that a chemical reaction has taken place. As molecules of a gas are held by Vander waal forces so, this gas will readily escape into the atmosphere.

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3 years ago
John dalton postulated several statements about the atom. which statement did he NOT state?
algol [13]

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7 0
3 years ago
What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.
pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

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ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

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mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in         184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in   4.867 g of MgBr₂ we have Y g of Br⁻

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4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

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