Vinegar is a solution of acetic acid in water. If certain vinegar has a concentration of 1.5% v / v.

What percentage of pure alcohol is in a shot of 100 proof liquor?

Alex_Xolod [135]

<span>First of all, an ounce of hard liquor (gin, vodka, rum, whiskey, or scotch) contains 64 calories for 80 proof varieties, and 80 calories for those that are100 proof. The proof refers to the percentage of alcohol in the liquor (80 proof is 40 percent alcohol; 100 proof is 50 percent alcohol)</span>

7 0

4 years ago

Which of the following statements are true of thermal energy and kinetic

Karolina [17]

Answer:

A. All the molecules or atoms in motion have kinetic energy.

B. All the molecules or atoms in motion have thermal energy.

C. Each molecule or atom in motion has thermal energy.

D. Each molecule or atom in motion has kinetic energy.

Explanation:

6 0

2 years ago

Any mixture that is heterogeneous on a microscopic level is a

Nadya [2.5K]

Solution is the answer.

5 0

4 years ago

Write the chemical equation for the reaction of lithium hydroxide (LiOH) with an Arrhenius acid of your choosing. Name and write

Inessa05 [86]

<h3>Answer:</h3>

LiOH(aq) +HCl(aq)→ LiCl(aq) + H₂O(l)

Salt formed is LiCl

<h3>Explanation:</h3>

Arrhenius acid refers to a substance that ionizes in water to generate protons or hydrogen ions.
Examples of Arrhenius acid include acids such as HCl, H₂SO₄ and HNO₃.
A reaction between Lithium hydroxide and an Arrhenius acid such as HCl will yield a salt and water.

That is;

LiOH(aq) +HCl(aq)→ LiCl(aq) + H₂O(l)

The salt formed is LiCl
The reaction is an example of neutralization reaction.

5 0

4 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago