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1 year ago

(b) A student states, "When the alcohol sample was at a temperature of 500 K, all the particles

1 answer:

7 0

The kinetic energy of particles depends on temperature. As temperature increases, the kinetic energy of particles increases.

Temperature is defined as a measure of the average kinetic energy of the molecules of a substance. The higher the temperature of a substance the greater its kinetic energy.

From the foregoing, when the temperature is increased from 400K to 500K, greater kinetic energy is imparted to the molecules of the substance hence they move faster according to the kinetic theory of matter.

Learn more: brainly.com/question/10079361

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How many solutions does the system have? zero one two three four infinitely many.

Ber [7]

Answer:

the system has infinitely many solutions.

Explanation:

The system is 2x + y = 1 and 4x + 2y = 2. Solutions to a system are the intersection points. Since these two lines are the same line they intersect everywhere. There are infinitely many solutions.

hope this helps

plz mark brainliest

4 0

2 years ago

Can you help me balance the equations from 1-10 please and thank you.

Korolek [52]

Answer:

2 NaBr + Ca(OH)2 → CaBr2 + 2NaOH
2 NH3 +H2SO4 → (NH4)2(SO4)

8 0

2 years ago

What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o

lyudmila [28]

Answer:

$\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%$

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

Thus, the atom percent turns out:

$\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%$

Best regards.

4 0

3 years ago

Can the pH scale be utilized for all acids (Arrhenius, Bronsted-Lowry, and Lewis)? Give examples of substances from each definit

CaHeK987 [17]

Answer:

Explanation:

The pH scale is a scale graduated from 0-14 which shows the degree of acidity of alkalinity of a substance. The pH scale is graduated in such a way that 0-6.9 indicates acidity, 7.0 indicate a neutral substance, while a pH of 8-14 indicates alkalinity respectively.

There are three main definitions of acids/bases

- Arrhenius definition

-Brownstead-Lowry definition

-Lewis definition

Arrhenius explains acids as any substance that produces hydrogen ions as its only positive ion in solution while a base produces hydroxide ions as its only negative ion in solution. The pH scale is based on corresponding values of pH derived from aqueous solutions of these substances.

However, not all acids/bases produces hydrogen or hydroxide ions in solution. Brownstead-Lowry definition of acids and Lewis definition of acids could be extended to nonaqueous media where the pH can not be measured as there are no hydrogen or hydroxide ions present in the solution.

This implies that pH measurement may not apply to acids/bases in the all the categories of acids/bases hence it can not be utilized for all acids and bases.

Arrhenius - sodium carbonate

Brownstead-Lowry - concentrated HF

Lewis acid - AlCl3

6 0

3 years ago

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

<u>Explanation:</u>

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

4 0

3 years ago