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PSYCHO15rus [73]
3 years ago
15

Fill the blanks :

Chemistry
1 answer:
andreev551 [17]3 years ago
5 0

Answer:

Hi, Friend!

My Answer would  be Number 3 Frequency

Explanation:

Frequency describes the number of waves that pass a fixed place in a given amount of time.

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VLD [36.1K]

Answer:

arthropod

Explanation:

hope this helps

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8 0
3 years ago
How do cm3 compare to mililiters
azamat
Cm^3=1cm*1cm*1cm
=10mm*10mm*10mm
=1000 mm^3
4 0
3 years ago
How to remember all your answers for a test?
Troyanec [42]

rewrite them and say them out loud or in your head repeatedly

8 0
4 years ago
How many atoms are in 44 grams of sulfur?
horrorfan [7]

Answer:

Explanation:

The Avagadro constant = 6.022 x 10²³ denotes the no of molecules in one mole of the substance

Regarding Sulphur Dioxide, the molar mass is 64.

Hence 44 g means 44/64 mole of SO2

So no of molecules in 44g of SO2 = 6.022 x 10²³ x 44/64

On simplification,

11/16 x 6.022 x 10²³ molecules are present in 44g of SO2

3 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
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