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mart [117]
3 years ago
13

g modenr vacuum pumps make it easy to attain pressures of the order of in the laboratory. at a preasusure of 6.75 atm and an ord

inary temperature of 290.0 k, how many molecules are present in a volume of 1.07 cm
Chemistry
1 answer:
Ratling [72]3 years ago
7 0

Answer:

Number of molecules = 1.8267×10^20

Explanation:

From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;

PV = nRT

where

P =pressure

V =volume

n = the number of moles

R is the gas constant equal to 0.0821 L·atm/mol·K

T is the absolute temperature

Given:

P = 6.75 atm;

T = 290.0 k,

; V = 1.07 cm³ = 0.001 L

( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)

n = 3.0335167*10^-4 moles

But there are 6.022×10²³ molecules in 1 mole,

Number of molecules = 1.8267×10^20

You might be interested in
Transition metals often form ions with a charge of _____-
Bess [88]
+1 for transition metals
5 0
3 years ago
Three different atoms or atomic cations with 3 electrons.
sveticcg [70]

Answer:

Li, Be^+, B^{2+}

Explanation:

To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).

The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).

Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, Be^+.

We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: B^{2+}.

6 0
3 years ago
A solution is made by mixing 37.g of thiophene C4H4S and 72.g of heptane C7H16. Calculate the mole fraction of thiophene in this
guajiro [1.7K]

Answer:

0.38

Explanation:

Molar mass of thiophene= 84g/mol

Mass of thiophene = 37g

Number of moles= 37/84= 0.44 moles

Molar mass of heptane= 100 g/mol

Mass of heptane = 72g

Number of moles = 72/100= 0.72 moles

Total number of moles= 0.44 + 0.72= 1.16 moles

mole fraction of thiophene = 0.44/1.16= 0.38

4 0
3 years ago
The standard enthalpy change for the following reaction is 873 kJ at 298 K.
Flura [38]

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

2KCl(s)\rightarrow 2K(s)+Cl_2(g)  \Delta H_1=873kJ

Now we have to determine the value of \Delta H for the following reaction i.e,

K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s) \Delta H_2=?

According to the Hess’s law, if we divide the reaction by half then the \Delta H will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=\frac{1}{2}\times (-873kJ)

\Delta H_2=-436.5kJ

Hence, the value of \Delta H_2 for the reaction is -436.5 kJ.

8 0
3 years ago
Why are parentheses used to write the formula AI(OH)3
Yakvenalex [24]

Because they are not on the periodic tsble they are on the back and they goes together

7 0
3 years ago
Read 2 more answers
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