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3 years ago

The definition to reactivity with vinegar

1 answer:

3 0

Reactivity to vinegar means that it is a substance that reacts to vinegar forming new pure substances. If a certain is labeled as such then it means that it will react to vinegar so it is not advisable to store them together cause they might contact with each other.

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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne

Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0

3 years ago

An atom of sodium can transfer its one valence electron to an atom of...

diamong [38]

Answer: Either give off or gain electrons to make a total of 8

8 0

2 years ago

1. A 25 g rock is placed in a graduated cylinder with water. The volume of the liquid rises from 18.3 mL to 21.4 mL Calculate th

MaRussiya [10]

Answer:

1. $\rho=8.06g/cm^3$

2. $H=10cm$

Explanation:

Hello,

1. In this case, since the volume of the rock is obtained via the difference between the volume of the cylinder with the water and the rock and the volume of the cylinder with the water only:

$V=21.4mL-18.3mL=3.1mL$

Thus, the density turns out:

$\rho=\frac{m}{V}=\frac{25g}{3.1cm^3} \\\\\rho=8.06g/cm^3$

2. In this case, given the density and mass of aluminum we can compute its volume as follows:

$V=\frac{m}{\rho}=\frac{1080g}{2.7g/cm^3}=400cm^3$

Moreover, as the volume is also defined in terms of width, height and length:

$V=W*H*L$

The height is computed to be:

$H=\frac{V}{L*W}=\frac{400cm^3}{5cm*8cm}\\ \\H=10cm$

Best regards.

3 0

3 years ago

A coffee cup calorimeter will not be used to directly measure the enthalpy of magnesium combustion because

yanalaym [24]

Answer:

A coffee cup calorimeter is great for measuring heat flow in a solution, but it can't be used for reactions that involve gases since they would escape from the cup. The coffee cup calorimeter can't be used for high-temperature reactions, either, because they would melt the cup.

8 0

3 years ago

Read 2 more answers

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c

Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

6 0

3 years ago