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3 years ago

Which property altered during a chemical change is not altered during a physical change?

2 answers:

7 0

<h3>During physical changes, the composition od the original substance is not altered, but the properties of the original substance are altered. During a chemical change the composition od the original substance is not altered and the change is irreversible. Melting of butter and wax is an example of chemical changes.</h3>

Mashutka [201]3 years ago

7 0

The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition.

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Does the model below represent a mixture or a substance ?

Vikentia [17]

Answer:

substance

Explanation:

A mixture is when two or more <u>different</u> atoms/molecules are together, but not joined.

A substance is when the <u>same </u>atom/molecule is in a group together.

In this example, it is a substance because it is comprised of the same molecule not joined all together. If you wanted a mixture, other colored atoms/molecule (e.g. add green atoms) would change it to this property.

4 0

3 years ago

Did they scientist believe in God

Nataliya [291]

Answer:

some of them not believe in God

6 0

3 years ago

The half life of an isotope X is 3 days how long will it take for 1/2 of a 10 G sample to decay

lorasvet [3.4K]

Answer: It will take 3 days for half of a 10 g sample to decay.

Explanation:

Half-life of sample of an isotope X = 3 days

$\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{3 days}=0.231 day^{-1}$

$N=N_o\times e^{-\lambda t}$

$\lnN=-\lambda t\lnN_o$

Sample decayed = $\frac{N_o}{2}=\frac{10 g}{2 g}=5 g$

N= $N^o-\text{sample decayed}=10 g-5 g=5 g$ , time = t

$\ln[5 g]=-0.231 day^{-1}\times t\ln[10]$

$\ln[\frac{5}{10}]=-0.231\times t$

$\ln\frac{1}{2}=-0.693=-0.231\times t$

t = 3 days

It will take 3 days for half of a 10 g sample to decay.

3 0

3 years ago

How are real Gases Different from ideal Gases?

madam [21]

Answer:

Explanation

Explanation:

Real gases have small attractive and repulsive forces between particles and ideal gases do not. Real gas particles have a volume and ideal gas particles do not. Real gas particles collide in-elastically (loses energy with collisions) and ideal gas particles collide elastically.

8 0

4 years ago

calculate the tonnes of co2 in the atmosphere. use the current co2 concentration of 405 ppm. Answer as tonnes as carbon.

Alex Ar [27]

Answer:

2,0894x10¹² tonnes of CO₂ in atmosphere

Explanation:

The total mean mass of the atmosphere is 5.1480×10¹⁸ kg. CO₂ is an important race gas in Earth's atmosphere because it is part of the carbon cycle.

405 ppm of CO₂ in atmosphere means 405 mg of CO₂ per kg of atmosphere. If the mass of the atmosphere is 5.1480×10¹⁸ kg:

$\frac{405mg CO_2}{1kg atmosphere} * 5,1480x10^{18} kg Atmosphere = 2,0849x10^{21}mg CO_2$

In tonnes:

$2,0849x10^{21}mg CO_2 * \frac{1kg}{1x10^{6} mg} * \frac{1 tonne}{1000 kg} =$ <em>2,0894x10¹² tonnes of CO₂ in atmosphere</em>

I hope it helps!

7 0

3 years ago