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Strike441 [17]
3 years ago
6

Is color change a physical or chemical change?

Chemistry
1 answer:
bagirrra123 [75]3 years ago
3 0
It's a chemical change because it's changing the object's property. 
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gladu [14]

Answer

Tq so much.

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5 0
3 years ago
Read 2 more answers
Mass defect for oxygen-16 = 0. 13261 amu How many kilograms does the mass defect represent? 1. 66 × 10-27 kg 2. 20 × 10 -28 kg 3
Zepler [3.9K]

Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.

<h3>What is mass defect?</h3>

Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.

We know that, 1 amu = 1.6 × 10⁻²⁷ kg

Given that, mass defect for oxygen-16 = 0.13261 amu

To calculate this defect in terms of kilograms, we have to convert into kg unit as:

0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu

0.13261 amu = 2.20 × 10⁻²⁸ kg

Hence option (2) is correct.

To know more about Mass defect, visit the below link:

brainly.com/question/4334375

7 0
2 years ago
The density of water at 40°c is 0.992 g/ml. what is the volume of 3.45 g of water at this temperature?
wariber [46]
3.45÷0.992=3.48ml
...
8 0
3 years ago
Use the table to calculate ΔH°rnx for the following reaction
alexdok [17]

Answer:

The answer to your question is ΔH° rxn = -1343.9 kJ/mol

Explanation:

                       P₄O₆ (s)  +  2 O₂ (g)   ⇒  P₄O₁₀

ΔH°rxn = ?

Formula

ΔH°rxn = ∑H° products - ∑H° reactants

H° P₄O₆ = -1640.1 kJ/mol

H° O₂ = 0 kJ/mol

H° P₄O₁₀ = -2984 kJ/mol

-Substitution

ΔH° rxn = (-2984) - (-1640.1) - (0)

-Simplification

ΔH° rxn = -2984 + 1640.1

ΔH° rxn = -1343.9 kJ/mol

7 0
3 years ago
It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water. Two solutions are available, the
bekas [8.4K]

Answer:

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

Explanation:

Let the mass of the first solution be x and second solution be y.

Amount solution required = 1250 kg

x + y = 1250 kg....[1]

Percentage of ethanol in required solution = 12% of 1250 kg

Percentage of ethanol in solution-1 = 5% of x

Percentage of ethanol in required solution = 25% of y

5% of x +  25% of y =12% of 1250 kg

\frac{5}{100}\times x+\frac{25}{100}y=\frac{12}{100}\times 1250 kg

x + 5y = 3000 kg...[2]

Solving [1] and [2] we :

x = 437.5 kg   , y =  812.5 kg

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

8 0
3 years ago
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