Ask question

Ask question

All categories

3 years ago

8

Calculate the mass, in grams, of each sample. a. 1.1 * 1023 gold atoms b. 2.82 * 1022 helium atoms c. 1.8 * 1023 lead atoms d. 7

.9 * 1021 uranium atoms

1 answer:

Vlada [557]3 years ago

5 0

Answer:blablablablablabla

Explanation:

Blablabla

You might be interested in

Hydrates that have a low vapor pressure and remove moisture from air are said to be ___. Question 8 options: effloresce hygrosco

-Dominant- [34]

Answer:

Hygroscopic

Explanation:

An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.

Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c

Efflorescence compounds gives off their water of crystallization to the atmosphere.

4 0

3 years ago

1. What happens when like charges are brought closer to each other?<br><br><br>

deff fn [24]

Like charges repel each other; unlike charges attract. Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges. The size of the force varies inversely as the square of the distance between the two charges.

5 0

3 years ago

Read 2 more answers

How many variables are there in science

Sedaia [141]

There are three variables independent, dependent ,and controlled

7 0

3 years ago

Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce

kkurt [141]

Explanation:

The given data is as follows.

P = 3 atm

= $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$

= $3.03975 \times 10^{5} Pa$

$V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$ (as 1 L = 0.001 $m^{3}$ ),

$V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

W = $P \times \Delta V$

or, W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

W = $P \times (V_{2} - V_{1})$

= $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

= 1823.85 Nm

or, = 1823.85 J

As internal energy of the gas $\Delta E$ is as follows.

$\Delta E$ = Q - W

= 800 J - 1823.85 J

= -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0

3 years ago

What would be the new volume if the temperature on 800 mL of gas is changed

dolphi86 [110]

I think you divide something

8 0

3 years ago