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3 years ago

What evidence of a chemical reaction might you see in the following cases?

1 answer:

5 0

For bleaching a stain, there will be an identifiable odor and it will easily absorb colors and/or stains. Burning a match also has odor involved, but the smoke that is released is from a hot fire that is chemically produced by the formula to make the tip of the match.

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Black and green tea have a growing body of research suggesting there are positive health benefits on because of the anti-oxidant

andrezito [222]

Keeping in mind a total ignorance of both the health benefits of these teas, and the interaction between milk and antioxidants, I believe that it is possible that milk could hinder these benefits.

Tea is usually a hot beverage. Milk, when added to this beverage, would easily dissolve. When a solute (milk) dissolves in a solvent (tea), the chemical properties of the resulting solution can become quite distinct from both of the original substances. It seems possible that the same chemical properties of tea that make it healthy could be altered by the addition of milk.

3 0

3 years ago

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

2 years ago

What observation about light supported Einstein's theory?

Llana [10]

In 1905 Albert Einstein had proposed a solution to the problem of observations made on the behaviour of light having characteristics of both wave and particle theory. From work of Plank on emission of light from hot bodies, Einstein suggested that light is composed of tiny particles called photons, and each photon has energy.

Light theory branches in to the physics of quantum mechanics, which was conceptualised in the twentieth century. Quantum mechanics deals with behaviour of nature on the atomic scale or smaller.

As a result of quantum mechanics, this gave the proof of the dual nature of light and therefore not a contradiction.

5 0

3 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:

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

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

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

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

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (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

What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in

Viktor [21]

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

3 0

2 years ago