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Airida [17]
3 years ago
12

I have this chemistry question and I’m really struggling to answer it. Can someone please help me figure out the steps & the

solution I need to find this answer?

Chemistry
1 answer:
Digiron [165]3 years ago
7 0

Answer: 133,333.333 meters.

Explanation:

I forgot how to explain this looool.

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3 0
3 years ago
State the postulate of Bohr theory
KengaRu [80]

Answer:

Bohr's model of the hydrogen atom is based on three postulates:

1) An electron moves around the nucleus in a circular orbit,

2) An electron's angular momentum in the orbit is quantised,

3) The change in an electron's energy as it makes a quantum jump from one orbit to another is always accompanied by the emission or absorption of a photon. Bohr's model is semi-classical because it combines the classical concept of electron orbit (postulate 1) with the new concept of quantisation ( postulates 2 and ).

4 0
3 years ago
How many molecules are in 3.6 grams of NaCl?
raketka [301]

Answer:

\boxed{\text{None}}

Explanation:

There are no molecules in NaCl, because it consists only of ions.

However, we can calculate the number of formula units (FU) of NaCl.

Step 1. Calculate the moles of NaCl

\text{No. of moles}=\text{3.6 g NaCl}\times \dfrac{\text{1 mol NaCl}}{\text{63.54 g NaCl}} = \text{0.0567 mol NaCl}

Step 2. Convert moles to formula units

\text{FU} = \text{0.0567 mol NaCl} \times \dfrac{6.022 \times 10^{23}\text{ FU NaCl}}{\text{1 mol NaCl}}\\\\= 3.4 \times 10^{22} \text{ FU NaCl}

There are \boxed{3.4 \times 10^{22} \text{ FU NaCl}} in 3.6 g of NaCl.

6 0
3 years ago
Which particle is emitted when an atom of 85Kr spontaneously decays?
stiv31 [10]
A Beta particles is emitted when an atom of 85Kr spontaneously decays.
5 0
3 years ago
Read 2 more answers
The nonvolatile, nonelectrolyte DDT , C14H9Cl5 (354.5 g/mol), is soluble in benzene, C6H6. Calculate the osmotic pressure (in at
Ainat [17]

Step 1

The osmotic pressure is calculated as follows:

\begin{gathered} \pi\text{ = C x R x T} \\ C\text{ = molarity = }\frac{moles\text{ of solute}}{Volume\text{ of solution \lparen L\rparen}} \\ R\text{ = 0.082 }\frac{atmxL}{mol\text{ x K}} \end{gathered}

-------------

Step 2

<em>Information provided:</em>

The mass of solute = 13.6 g

Volume of solution = 251 mL

Absolute temperature = T = 298 K

The molar mass of solute = M = 354.5 g/mol

-------------

Step 3

Procedure:

1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L

---

C = moles of solute/volume of solution (L)

C = mass of solute/(molar mass x Volume (L))

C = 13.6 g/(354.5 g/mol x 0.251 L)

C = 0.153 mol/L

---

π = C x R x T

π = 0.153 mol/L x 0.082 atm L/mol K x 298 K

π = 3.74 atm

Answer: π = 3.74 atm

4 0
1 year ago
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