Ask question

Ask question

All categories

3 years ago

7

Which lists mixtures, in order, from the smallest particles to the largest particles?

1 answer:

Kisachek [45]3 years ago

4 0

Answer:

A. Solution, Colloid, Suspension

Hope i could help

You might be interested in

Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(

MakcuM [25]

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

The magnitude of the angular momentum is $L = 6\sqrt{\frac{h}{2 \pi} }$

The generally formula for Orbital angular momentum is mathematically represented as

$L = \sqrt{(l * (l + 1)) } * \frac{h}{2 \pi}$

Where $l$ is the quantum number

now

We can look at the given angular momentum in this form as

$L = 6\sqrt{\frac{h}{2 \pi} } = \sqrt{36} * \sqrt{\frac{h}{2 \pi}} }$

comparing this equation to the generally equation for Orbital angular momentum

We see that there is no quantum number that would satisfy this equation

5 0

3 years ago

Determining Density and Using Density to Determine Volume or Mass

Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = $\frac{mass}{volume}$

ρ=m/v ,m=ρv, v=m/ρ

(a)m=1*10g , v=7.36cm³

ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

ρ=374.5/41.8 =8.96g/cm³

mass of ethanol=15g, density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0

3 years ago

CsBr formula name???

olganol [36]

Answer

PubChem CID/molecular formula

Explanation:

Cesium bromide

PubChem CID 24592

Molecular Formula CsBr or BrCs

Synonyms CESIUM BROMIDE 7787-69-1 Caesium bromide Cesiumbromide Cesium bromide (CsBr) More...

Molecular Weight 212.81 g/mol

Component Compounds CID 260 (Hydrogen bromide) CID 5354618 (Cesium)

have a good day /night

may i please have a branllist

4 0

2 years ago

What is the formula for dihydrogen tetraoxide

Veseljchak [2.6K]

It's been a while... but I'm <em>pretty</em> sure it's

H₂O₄

3 0

3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago