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german
3 years ago
9

Why does magnesium chloride have a much higher melting point than hydrogen chloride ?

Chemistry
2 answers:
ra1l [238]3 years ago
6 0
The other chlorides are simple covalent molecules. Melting and boiling points: Sodium and magnesium chlorides are solids with high melting and boiling points because of the large amount of heat which is needed to break the strong ionic attractions. The rest are liquids or low melting point solids.Sep 16, 2015
pshichka [43]3 years ago
5 0
Because chloride is more reactive than hydrogen.
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Which unit can be used to express the rate of a reaction?
Ugo [173]
<span>A) mL / s

This is the amount of milliliters per second</span>
4 0
3 years ago
Read 2 more answers
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of
In-s [12.5K]

Answer:

The answer to your question is: 101.2 g of CO2

Explanation:

C = 27.6 g

O₂ = 86.5 g   remained 12.9 g

O₂ that reacted = 86.5 - 12.9 = 73.6 g

                     C     + O₂      ⇒        CO₂      The equation is balanced

                    27.6    73.6                 ?

MW               12        32                  44

Rule of three

                        12 g of C------------------  44 g  CO2

                       27.6 g C  ------------------    x

                  x = 27.6(44)/12 = 101.2 g of CO2

                       32 g of O2 ---------------    44 g of CO2

                        73.6 g of O2 ------------      x

                  x = 73.6(44)/32 = 101.2 g of CO2

6 0
3 years ago
por favor es urgente!!!!!!!!.... 1 -¿cuántos átomos- gramos de cloro hay en 88,75 g de dicho elemento? 2-¿calcular los moles de
Mamont248 [21]

Answer:

don't understand your language

6 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
The radioactivity of U-235 is of low intensity. Why then were the people of Hiroshima exposed to high intensity radiation in the
Marina86 [1]

Answer:

Explanation:

What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.

7 0
3 years ago
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