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3 years ago

What is the weakness of particle theory?

Chemistry

1 answer:

stich3 [128]3 years ago

5 0

Answer:

The particle model does not take into account: the size and shape of particles. the space between particles.

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What types of intermolecular forces are found in CH<br> 4

Sphinxa [80]

London dispersion forces

8 0

3 years ago

Use the periodic table to identify the molar mass of each element below. Answer without doing any calculations.

Delicious77 [7]

Answer:

Beryllium (Be) : 9.01 g/mol

Silicon (Si) : 28.09 g/mol

Calcium (Ca) : 40.08 g/mol

Rhodium (Rh) : 102.91 g/mol

Explanation:

8 0

3 years ago

Read 2 more answers

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

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

8 0

3 years ago

I need to know the following above

arsen [322]

3Zn + 8HNO3 ---> 3Zn(NO3)2 + 4H2O + 2NO IF IT IS COLD AND DILUT NITRIC ACID .

IF IT IS HOT AND CONCENTRATED THEN:

Zn+ 4HNO3 ---> Zn(NO3)2 +2H2O +2NO2

6 0

3 years ago

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$