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9966 [12]
3 years ago
8

In one experiment a student accidentally injected a syringe filled with 10 mL of 12 M H2SO4 into the pressure vessel. The metal

reacted, H2 was produced and the final pressure was recorded. Can this experiment be used to determine the identity of the unknown metal, or is the data unusable?
Yes or No
Chemistry
1 answer:
Volgvan3 years ago
4 0
Yes i think it’s the correct answer !
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Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissol
Stells [14]

Here is the complete question.

Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol (CH3CH2OH; density = 0.7893 g/mol). Round to four significant digits

Answer:

0.9567 mol

Explanation:

Given that:

mass of glycerol = 1.61 g

molar mass of glycerol = 92.1 g/mol

no of mole = \frac{mass}{molar mass}

∴ number of moles of glycerol (n_{glycerol}) = \frac{1.61}{92.1}

= 0.0175 mol

Volume of ethanol = 22.60 mL

Density of ethanol = 0.7893 g/mL

Since Density = \frac{mass}{volume}

∴  mass of ethanol = density of ethanol × volume of ethanol

mass of ethanol =  0.7893 g/mL × 22.60 mL

mass of ethanol =  17.838 g

Number of moles of ethanol (n_{ethanol}) = \frac{17.838}{46.0684}

= 0.387 mole

∴ the mole fraction of the solvent can be determined as:

X_{solvent} = X_{ethanol} = \frac{n_{ethanol}}{n_{glycerol}+n_{ethanol}}

=\frac{0.387 mol}{(0.0175+0.387)mol}

= \frac{0.387mol}{0.4045mol}

= 0.95673671199

≅ 0.9567 mol

∴ The mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol is = 0.9567 mol

7 0
3 years ago
How many moles of carbon dioxide can be produced when 3.05 of calcium carbonate are heated​
vredina [299]

Answer:

0.0305mol

Explanation:

CaCO3 ---> CaO + CO2

Mass of CaCO3 mol = 40 + 12 + (16 x 3) = 100g/mol

Number of CaCO3 moles heated = 3.05/100 = 0.0305 mol

One CaCO3 mol produces 1 mol CO2

Therefore 0.0305mol  of CO2 produced.

3 0
2 years ago
Explain why lithium fluoride is a solid at room temperature
SashulF [63]
Lithium fluoride is a solid at room temperature because it is a salt that is held together by ionic bonds. Lithium. fluoride has a giant ionic structure.
8 0
3 years ago
How many milliliters of a 0.211 M HI solution are needed to reduce 24.0 mL of a 0.354 M KMnO4 solution according to the followin
Novay_Z [31]

Answer:

The answer to your question is 242 ml

Explanation:

Data

HI 0.211 M   Volume = x

KMnO₄ 0.354 M   Volume = 24 ml

Balanced Chemical reaction

     12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O

Process

1.- Calculate the moles of KMnO₄  0.354 M in 24 ml

Molarity = moles / volume (L)

moles = Molarity x volume (L)

moles = 0.354 x 0.024

moles = 0.0085

2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,

              12 moles of HI --------------- 2 moles of KMnO₄

                x                     --------------- 0.0085 moles of KMnO₄

             x = (0.0085 x 12)/2

             x = 0.051 moles of HI

3.- Calculate the milliliters of HI 0.211 M

Molarity = moles/volume

Volume = moles/molarity

Volume = 0.051/0.211

Volume = 0.242 L or Volume = 242 ml

8 0
3 years ago
Use the following equation to answer the questions below:
Gala2k [10]

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass  *  Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

3 0
3 years ago
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