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3 years ago

What did Rutherford observe that surprised him?

Chemistry

2 answers:

Valentin [98]3 years ago

8 0

Answer:

Some of the Alpha particles shot at the gold foil bounced back.

Explanation:

lana [24]3 years ago

7 0

Answer:

Some of the Alpha particles shot at the gold foil bounced back.

Explanation:

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How many

Aleksandr-060686 [28]

Answer:

A. 1.50

Explanation:

9.01 x 1023 molecules* 1 mol/ 6.022×1023 molecules= 1.49618067087 moles

Rounded would be 1.50

5 0

3 years ago

How many grams of carbonic acid were produced by the 3.00g sample of NaHCO3

Alexxx [7]

I dont know what subject is this

8 0

4 years ago

2ch4(g) c2h2(g) 3h2(g) describe what is happening within the system when it is at equilibrium in terms of concentrations, reacti

SashulF [63]

Balanced chemical reaction: 2CH₄(g) ⇄ C₂H₂(g) + 3H₂(g).

1) In a chemical reaction, chemical equilibrium is the state in which both reactants (methane CH₄) and products (ethyne C₂H₂ and hydrogen H₂) are present in concentrations which have no further tendency to change with time.

2) At equilibrium, both the forward and reverse reactions are still occurring.

3) Reaction rates of the forward and backward reactions are equal and there are no changes in the concentrations of the reactants and products.

4 0

3 years ago

Read 2 more answers

In the titration of acetic acid, CH3COOH, a 0.20 M solution of NaOH is used. A volume of 15.00 mL of CH3COOH solution is

yan [13]

Answer:

naci

Explanation:

6 0

3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago