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3 years ago

6

A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes

t demonstration that she could use?

1 answer:

Jlenok [28]3 years ago

5 0

when you crumble paper is still paper when you burn down paper you get ashes. that is want is call chemical and physical changes.

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A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

Yakvenalex [24]

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

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2 years ago

Calculate the mass percent composition for Al(HCO3)3

ZanzabumX [31]

Answer:

salamat sapoints po lods thaks

7 0

2 years ago

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave

timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

$n_i=8$

$\lambda=3745\ nm$

The amount of energy change during the transition is given by :

$\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

And

$\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

Plugging all the values we get :

$\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5$

So, the final level of the electron is 5.

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3 years ago

Negatively charged particles within the atom are called what?

Ivahew [28]

Negatively charged particles in atom - Electrons

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3 years ago

____________ is a colorless, odorless gas that forms when fuels are not completely burned.

-Dominant- [34]

Your answer is Carbon monoxide

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3 years ago

Read 2 more answers