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2 years ago

6

Identify the type of bonding, molecular geometry (shape), and intermolecular forces experienced by the compounds HF and HBr. Whi

ch substance would experience a stronger attraction between its molecules, and why?

1 answer:

Anestetic [448]2 years ago

8 0

Im not sure but i think its hbr. i hope im right and i hope this helped,

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Ca(OH)2 + H3PO4 = Ca3(PO4)2 + H2O

larisa [96]

Make a quick chart with each element represented, and count them up. HINT - leave the polyatomic anions together - in this case, PO4

Left Right
1 Ca 3
2 O 1
5 H 2
1 PO4 2

Begin by balancing like finding common denominators of fractions - apply to both sides:
I started by adding a 2 in front of H3PO4 on the left, them 6 in front of H2O on the right. Last, a 3 in front of Ca (OH)2. Then, re-count using the chart format to make sure you're right.

3Ca(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O

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2 years ago

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Daniel [21]

Answer:

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2 years ago

How many O2 molecules react with 47 CH4 molecules according to the preceding equation CH4 + O2 → CO2 + H2O

natka813 [3]

I think 93.71 mol.
You need to use stochio

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2 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

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3 years ago

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Iron + sulphur - iron sulphide

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3 years ago

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