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Aleks [24]
2 years ago
11

When the metallic element sodium combines with the nonmetallic element bromine (Br2), is the product a solid, liquid, or gas at

room temperature?
I will give brainliest if you can explain how one would know this information.
Chemistry
1 answer:
Luden [163]2 years ago
4 0

Answer: it is a solid at room temperature.

Explanation:

The ions of the two reactants are Na{+} and Br{-}, so the formula of the product is NaBr.  

and when the NaCl is a solid a room temperature. and when NaBr has similar properties to NaCl, so NaBr is a solid at room temperature too.

so our chemical formula for the product is NaBr and It's a solid at room temperature.

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Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
3 years ago
How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex
Brut [27]
When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M 
so by substitution we can get [Ca2+] 
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
                                                      = 0.00167 mol

4 0
3 years ago
What is the mass of one mole of H2o
puteri [66]

Answer:the answer is 18.01528

Explanation:

5 0
3 years ago
Which type of front occurs when cold air and warm air are next to each other, but are at a standstill?
noname [10]
Occluded font. For sure
8 0
3 years ago
Breeder reactors are used to convert the nonfissionable nuclide U-238 to a fissionable product. Neutron capture by the U-238 is
m_a_m_a [10]

Answer:

Pu-239

Explanation:

Beta decay moves the element which undergoes the decay one place to the right in the periodic table since to conserve charge and being beta radiations an  electron we convert a neutron into a proton and an electron. In neutron capture we increase the atomic mas by one unit. We that in mind, lets solve the question:

U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 +  ₋₁⁰β ⇒ Pu-239 +  ₋₁⁰β

7 0
3 years ago
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