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andrezito [222]
4 years ago
6

What type of energy is this?

Chemistry
1 answer:
Firlakuza [10]4 years ago
6 0
2 correct answers: Chemical and neither endothermic or exothermic
You might be interested in
Which of the following reactions is a double displacement reaction? A) HCI (9) - Hz (g) + Cl2 (9) B) HCI (aq) + NaOH (aq) - H2O
Marianna [84]

Answer:

HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

Explanation:

A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.

5 0
3 years ago
What is the chemical name for LiC2H3O2 or C2H3O2Li
bixtya [17]

Answer:

AWAWDWDAD

Explanation:

3 0
3 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
4 years ago
16.0 grams of oxygen gas reacted with 80.0 grams nitrogen monoxide gas producing 25.0 grams of nitrogen dioxide gas in the lab.
Anon25 [30]

Answer:

Limiting reactant  = O₂

Excess reactant = NO

Theoretical yield of NO₂ = 46 g

Mass of excess reactant = 30 g

<h3 />

Explanation:

O₂ + 2NO → 2NO₂

Mole ratio for the reaction is;

1 : 2 → 2

mass of O₂ = 16 g

mass of NO = 80 g

mass of NO₂ = 25 g

molecular weight  of O₂ = 32 g/mol

molecular weight  of NO = 30 g/mol

molecular weight  of NO₂ = 46 g/mol

molar mass of O₂ = mass ÷ molecular weight = 16 g ÷ 32 g/mol = 0.5 mol

molar mass of NO = mass ÷ molecular weight = 80 g ÷ 30 g/mol = 2.67 mol

Since, 1 mole of O₂ requires 2 moles of NO for the combustion reaction, 0.5 mole shall require 1 mole of NO for the reaction. Thus, O₂ is the limiting reactant and NO is the excess reactant as it has an excess of 2.67 mol - 1 mol = 1.67 mol.

<h3>Theoretical yield of NO₂</h3><h3 />

1 mole of O₂ shall yield 2 moles of NO₂

Thus, 0.5 mole of O₂ shall yield 1 mole of NO₂

mass of NO₂ = molecular weight * molar mass = 46 g/mol * 1 mole = 46 g

<h3>Mass of Excess Reactant</h3>

1 mole of O₂ shall react with 2 moles of NO

Thus, 0.5 mole of O₂ shall yield 1 mole of NO

mass of NO = molecular weight * molar mass = 30 g/mol * 1 mole = 30 g

5 0
3 years ago
Select True or False: A mixture made from 10 mL of 1 M HCl and 20 mL of 1 M CH3COONa would be classified as a buffer solution.
DIA [1.3K]

Answer:

True.

Explanation:

There are three ways to make a buffer.

a.  Generating a solution of a weak acid and its conjugate base.

b. Adding to a solution of a weak acid a certain quantity of

strong base, so that the acid remains in excess.

c. adding a limited amount to a conjugate base solution

of strong acid so that the base remains in excess.

We are in c, in this situation.

How do you calculate, pH? We apply Henderson Hasselbach.

pH = pka + log (mmoles base - mmoles acid)/ mmoles acid

pH = pKa + log ((20 ml . 1 M - 1 ml . 10M) / 10 mmoles

pH = 4.76 + log 1 → 4.76 ⇒ pH = pKa

8 0
3 years ago
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