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Ivanshal [37]
3 years ago
10

A reaction at evolves of boron trifluoride gas.Calculate the volume of boron trifluoride gas that is collected. You can assume t

he pressure in the room is exactly . Round your answer to significant digits.
Chemistry
1 answer:
vampirchik [111]3 years ago
3 0

Answer:

3.10 L

Explanation:

<em>A reaction at 11.0°C evolves 133.mmol of boron trifluoride gas. Calculate the volume of boron trifluoride gas that is collected. You can assume the pressure in the room is exactly 1 atm. Round your answer to 3 significant digits.</em>

Step 1: Convert the temperature to the Kelvin scale

When working with gases, we must use the absolute temperature scale. We can convert the temperature from Celsius t Kelvin using the following expression.

K = °C + 273.15

K = 11.0°C + 273.15

K = 284.2 K

Step 2: Convert the amount of matter to moles

We will use the conversion factor 1 mol = 1000 mmol.

133 mmol × (1 mol/1000 mmol) = 0.133 mol

Step 3: Calculate the volume of boron trifluoride gas

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.133 mol × (0.0821 atm.L/mol.K) × 284.2 K / 1 atm

V = 3.10 L

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Answer:

Always equal to the total moles of the products.

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction equal to the mass of the reactants.

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After leaving the cockroaches in their metabolic chamber for one hour the gases in the chamber was measured. The fraction of oxy
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Reacting 35.4 ml of 0.220 m agno3 with 52.0 ml of 0.420 m k2cro4 results in what mass of solid formed
laila [671]
Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.

7 0
3 years ago
What can you infer about aspirin’s ester group?
Reil [10]

Answer:

Option B, aspirin’s ester group provides greater digestibility to aspirin

Explanation:

Aspirin ester group has three parts

  1. carboxylic acid functional group (R-COOH)
  2. ester functional group (R-O-CO-R')
  3. aromatic group (benzene ring)

Aspirin is a weak acid and hence it cannot dissolve in water readily. The reaction of Aspirin ester group with water is as follows -

aspirin

(acetylsalicylic acid) + water → salicylic acid + acetic acid

(ethanoic acid)

Aspirin passes through the stomach and remains unchanged until it reaches the intestine where it hydrolyses ester to form the active compound.

8 0
3 years ago
Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di
pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)

Moles of calcium nitrate = \frac{31.3 g}{164 g/mol}=0.1908 mol

Moles of ammonium fluoride = \frac{38.7 g}{37 g/mol}=1.046 mol

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

\frac{1}{2}\times 1.046 mol=0.523 mol calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

\frac{2}{1}\times 0.1908 mol=0.3816 molof dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0
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