All categories

3 years ago

Name one object that would be considered a solid. Describe how the atoms move in a solid object.

Chemistry

1 answer:

Roman55 [17]3 years ago

7 0

Objects considered a solid:

Pencil

Water bottle

Phone

Table

Plant

Atoms are very attracted to one another in solid objects. They vibrate, but remain in fixed positions without moving.

You might be interested in

A molecule with a triple covalent bond is _____?

Contact [7]

A molecule with a triple covalent bond is N2
Na would be the best conductor of electricity

3 0

4 years ago

10. At 573K, NO2(g) decomposes forming NO and O2. The decomposition reaction is second order in NO2 with a rate constant of 1.1

leva [86]

Answer:

48.67 seconds

Explanation:

From;

1/[A] = kt + 1/[A]o

[A] = concentration at time t

t= time taken

k= rate constant

[A]o = initial concentration

Since [A] =[A]o - 0.75[A]o

[A] = 0.056 M - 0.042 M

[A] = 0.014 M

1/0.014 = (1.1t) + 1/0.056

71.4 - 17.86 = 1.1t

53.54 = 1.1t

t= 53.54/1.1

t= 48.67 seconds

Hence,it takes 48.67 seconds to decompose.

6 0

3 years ago

The mass of the products in a reaction is always _______ when compared to the mass of the reactants in a reaction. greater lower

BigorU [14]

Equal and balance

.......

6 0

3 years ago

Read 2 more answers

A) fluorite B) orthoclase C) apatite D) gypsum

Sidana [21]

C

I’m pretty sure it’s c

5 0

3 years ago

A sample of helium gas occupies 355ml at 23°c. if the container the he is in is expanded to 1.50 l at constant pressure, what is

ss7ja [257]

Answer: The final temperature would be 1250.7 K.

Explanation: We are given a sample of helium gas, the initial conditions are:

$V_{initial}=355mL=0.355L$ (Conversion factor: 1L = 1000 mL)

$T_{initial}=23\°C=296K$ (Conversion Factor: 1° C = 273 K)

The same gas is expanded at constant pressure, so the final conditions are:

$V_{initial}=1.50L$

$T_{initial}=?K$

To calculate the final temperature, we use Charles law, which states that the volume of the gas is directly proportional to the temperature at constant pressure.

$V\propto T$

$\frac{V_{initial}}{T_{initial}}=\frac{V_{final}}{T_{final}}$

Putting the values, in above equation, we get:

$\frac{0.355L}{296K}=\frac{1.50L}{T_{final}}$

$T_f=1250.7K$

5 0

3 years ago

Read 2 more answers