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polet [3.4K]
3 years ago
5

1. How would the series of figures change in the presence of a catalyst?

Chemistry
2 answers:
statuscvo [17]3 years ago
8 0

No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.

hope this explains it

If it does, please give it a brainliest :)))

Luda [366]3 years ago
8 0

Answer:

The figures represent a time sequence for a chemical reaction.

1. The presence of a catalyst would add new molecules, catalyst molecules, to the figures. Catalysts are not consumed as the reaction takes place, so its molecules would remain the same in all figures.

On the other hand, a catalyst speeds up a chemical reaction, so the quantity of reactant and product molecules would also change, with greater amounts of products and lower of reactants.

2. Yes, it would. As stated before, the presence of a catalyst would increase the amounts of products and decrease the amounts of reactants.

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if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
4 years ago
The density of silver is 10.5 g/cm3. what would be the mass (in grams) of a piece of silver that occupies a volume of 23.6 cm3?
Keith_Richards [23]
Hey there : !

density = 10.5 g/cm³

volume = 23.6 cm³

therefore:

D = m / V

10.5 = m / 23.6

m = 10.5 * 23.6

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hope this helps!
6 0
3 years ago
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Calculate the percent composition of muscovite mica. Its chemical
Vikki [24]

Answer:

Explanation:

(KF)2(Al2O3)3(SiO2)6(H2O)

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K₂F₂Al₆Si₆O₂₂H₂

2 x 39 + 2 x 19 + 6 x 27 + 6 x 28 + 22 x 16 + 2 x 2

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= 802

Percentage of K = 78 x 100 / 802

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= 20.2  %

Percentage of Si = 168 x 100 / 802

= 20.9  %

Percentage of O = 352 x 100 / 802

= 43.9 %

Percentage of H = 4 x 100 / 802

= .54  %

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