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3 years ago

I need to know this because I was absent and he was going and could not help me with it so I am on my own

Chemistry

2 answers:

valkas [14]3 years ago

6 0

Aw, I hope I can help.

1) C
(A meter stick is the only form of measurement that measures distance)

2) C
(other metric units all correspond with other forms of measurement)

3/4) I can’t measure the line physically from my phone but I think you’d be alright!

5) B
(a triple beam balance measures mass!)

6) B
(corresponds with triple beam balance)

7) A: 34.5g B: 104.3g C: 132.5g

8) C

9) A
(a graduated cylinder is the only form of measurement that measures volume)

10) A
(corresponds with a graduated cylinder!)

Brrunno [24]3 years ago

5 0

Triple Beam balance: used to weigh small things in grams
Meter stick: how long cm
Water Displacement: when handling liquids in a bottle sometimes it might not line up to line but go up at sides. measure from there.

1.C
2.C
3&4.metric stick to measure
5.b
6.g
7a.30g
7b.100g
7c.130g
8.A
9.A
10.A

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Minchanka [31]

I’m pretty the answer would be continental slope. :)
I really hope this helps.

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3 years ago

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The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

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3 years ago

What are in the energy levels that surround the nucleus of an atom?

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Electrons are the thing in the outer energy levels

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3 years ago

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Xelga [282]

Answer : For the experiment to study the diffusion across the semi permeable membrane, following are the safety steps that will help to handle chemicals safety;

Use droppers to transfer small amounts of chemicals into another glassware, it will prevent the unnecessary spillage of the chemical.
Use tongs to move hot objects as they may cause severe burns.
Check the test tubes for chips prior to using them otherwise it may leak the chemical inside it
Washing your hands after you finishing the lab experiment, this will ensure that food is not contaminated while consuming, with the chemicals that were stuck on the hands.

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3 years ago

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When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to c

Ber [7]

Taking into account definition of percent yield, the percent yield for the reaction is 57.08%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + Cl₂ → CH₃Cl + HCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
Cl₂: 1 mole
CH₃Cl: 1 mole
HCl: 1 mole

The molar mass of the compounds is:

CH₄: 16 g/mole
Cl₂: 70.9 g/mole
CH₃Cl: 50.45 g/mole
HCl: 36.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
Cl₂: 1 mole ×70.9 g/mole= 70.9 grams
CH₃Cl: 1 mole ×50.45 g/mole= 50.45 grams
HCl: 1 mole ×36.45 g/mole= 36.45 grams

Mass of CH₃Cl formed

The following rule of three can be applied: if by reaction stoichiometry 16 grams of CH₄ form 50.45 grams of CH₃Cl, 25 grams of CH₄ form how much mass of CH₃Cl?

$mass of CH_{3} Cl=\frac{25 grams of CH_{4}x 50.45grams of CH₃Cl }{16 grams of CH_{4}}$

<u><em>mass of CH₃Cl= 78.83 grams</em></u>

Then, 78.83 grams of CH₃Cl can be produced from 25 grams of CH₄.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield}x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know: