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Llana [10]
3 years ago
7

Suppose that you take a sample of 500 g of ocean water and let the water evaporate. The mass of the remaining salts is 17 g. Wha

t was the salinity of the ocean water
Chemistry
1 answer:
yanalaym [24]3 years ago
8 0

<u>Answer:</u> The salinity of ocean water is 34 ppt.

<u>Explanation:</u>

Salinity is defined as the number of grams of salts present per kilogram of seawater. It is expressed in parts per thousand.

The formula used to calculate the salinity of ocean water or sea water is:

\text{ Parts per thousand}=\frac{\text{Mass of salts}}{\text{Mass of sea water}}\times 1000

We are given:

Mass of salts = 17 g

Mass of sea water = 500 g

Putting values in above equation, we get:

\text{ Parts per thousand}=\frac{17g}{500g}\times 1000\\\\\text{ Parts per thousand}=34

Hence, the salinity of ocean water is 34 ppt.

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Heat energy is _____ when it moves from one room to another.
MArishka [77]
Hello,

Thanks for posting your question here on brainly.

There are 3 ways heat energy can move:<span> Radiation, conduction, and convection.
</span>
However, your answer to this is most likely conduction

hope this helps:)

please let me know if it's correct.



6 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
At 500 K the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) has Kp = 51 In an equilibrium mixture at 500 K, the partial pressure of NO is
Aleks [24]

Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }

Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)

7 0
3 years ago
When methane (CH4) gas is burned in the presence of oxygen, the following chemical reaction occurs.
liraira [26]
Idk try asking Crawford
4 0
3 years ago
Identify the species that was reduced in the titration reaction. Justify answer in terms of oxidation number 6h+(aq)+2mno4−(aq)+
egoroff_w [7]

Answer:

Explanation:

Reaction given

6 H⁺ + 2 MnO₄⁻ + 5 (COOH)₂ = 10CO₂ +8H₂O + 2 Mn⁺²

Oxidation number of Mn in MnO₄⁻

=  x - 4 x 2 = -1

x = 8 -1

+ 7

Oxidation no of Mn in Mn⁺² =  +2

So its oxidation no is decreased from + 7 to + 2  . Hence it is reduced.

5 0
3 years ago
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