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lilavasa [31]
3 years ago
10

Sketch a heating curve graph of temperature The b.p. of methanol is 65°C and that of

Chemistry
2 answers:
Reptile [31]3 years ago
8 0

Answer:

its 70*

Explanation:

noname [10]3 years ago
8 0
It’s 70* I belive it’s 70
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Three separate 3.5g blocks of al, cu, and fe at 25°c each absorb 0.505 kj of heat. which block reaches the highest temperature?
77julia77 [94]

The energy can be shown as:

Q = ms dT

Whereas, m is the mass of block

s is specific heat

dT is change in temperature.

Copper block having the lowest specific heat and thus having the higher change in temperature and therefore having the higher final temperature.

8 0
3 years ago
A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a
trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 +  2 HCl  -->  MnCl2  +  H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl)  X (0.0020 mole HCl / 1000ml) X (4.86 ml)    

= 4.86 x 10^-3 mole  

this is now dissolved in (70 + 4.86)  =  74.86 ml or 0.07486 L

thus [Mn(OH)2]  =  4.86 x 10^-3 mole / 0.07486 L  =  0.064921 M

Ksp =  [Mn2+][OH-]^2  =  4x^3  =  4(0.064921)^3  = 10.945 x 10^-4

6 0
3 years ago
Help with this question for chemistry!!
wel
Emperic formula is SO subscript 2. Molecular formula would be SO subscript 2 multiply 1 so it's the same answer.
7 0
3 years ago
Write the net ionic equation for the precipitation of cobalt(ii) carbonate from aqueous solution
andrey2020 [161]
Missing question: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium carbonate and cobalt(II) bromide are combined.<span>Balanced chemical reaction:
(NH</span>₄)₂CO₃(aq) + CoBr₂(aq) → CoCO₃(s) + 2NH₄Br(aq).
Net ionic reaction:
2NH₄⁺(aq) + CO²⁻(aq) + Co²⁺(aq) + 2Br⁻(aq) → CoCO₃ + 2NH₄(aq)+ 2Br(aq).
or CO²⁻(aq) + Co²⁺(aq) → CoCO₃(s).
4 0
3 years ago
Choose the atom with:
nata0808 [166]

Answer:

a )Li

b)O

c)F

Explanation:

a) Li-1s^2 2s^1

F-1s^2 2s^2 2p^5

it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy

b)oxygen ion is larger than Na because o have fewer proton

c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.

6 0
3 years ago
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