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Dahasolnce [82]
3 years ago
11

Change 23 degrees Celsius to F

Chemistry
2 answers:
Fed [463]3 years ago
8 0
Rewrite the formula C=5/9(F-32) substituting 23 for C: 23=5/9(F-32), then multiply both sides by the reciprocal of 5/9.
(9/5)*(23)=(9/5)*5/9(F-32)
41.4=F-32; add 32 to both sides.
41.4+32=F-32+32
73.4=F
Marat540 [252]3 years ago
7 0
It's 73.5 F cz c/5=f-32/9
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Nana76 [90]

Answer:

Therefore, when an atom becomes a positive ion is pulls its electrons closer, decreasing is radius moreover, when it becomes a negative ion, it pulls its electrons closer and decreases the radius.

5 0
2 years ago
Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi
uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

5 0
3 years ago
How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature? A. 51,3
expeople1 [14]
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:

E = m(ΔH)

where:
m = mass of substance
ΔH = latent heat of fusion or melting

According to data, the ΔH of mercury is approximately 11.6 kJ/kg.

E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J

The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
3 0
3 years ago
Read 2 more answers
Chapter 2.4.1. The ______ energy lost by the inward movement of nickel-iron within a differentiating molten planetary body would
Gwar [14]

Answer:b. gravitational, kinetic, thermal

Explanation:

The above explains the mechanism of the core forming process on earth/planet.

It is believed that this process might has contributed significantly to a planet's early stages heating. The time when these core-forming event happened is thought to have been mainly completed very early when Earth was young . The type of this event rather than it being seen as a single catastrophic event, it is likely to have been as a result of contractions on the earth severally.

The addition of partially differentiated material from another giant impact the rate of this spasm , and it increases each time the planet's mass is to increased.

This is a little on the history of planetary evolution.

8 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
4 years ago
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