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3 years ago

The density of a solid reactant affects the rate of a reaction. TRUE FALSE

Chemistry

2 answers:

levacccp [35]3 years ago

5 0

Answer:

The correct answer is "True".

Explanation:

The rate of a reaction is affected by multiple factors, including the density of a solid reactant. The higher the density of a solid reactant the higher the rate of a reaction will be. A higher density of reactant increases the probability that collisions take place, and the more collisions the more reactions will occur.

Shalnov [3]3 years ago

3 0

yes it is true it reàlly happens

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JOSE BOUGHT 750 BAGS OF PEANUTS FOR 375 HE INTENDS TO SELL EACH BAG FOR 0.15 MORE THAN HE PAID HOW MUCH SHOULD HE CHARGE FOR EAC

labwork [276]

He paid 2 dollars for each bag. Then add 15 cents for that and he would charge $2.15 for each bag.

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3 years ago

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Trans fats: Think about how the prefix trans- is used in naming alkenes.

wel

In an alkene, cis and trans isomers are possible because the double band is rigid, cannot rotate, has groups attached to the carbons of the double bond that are fixed relative to each other, and only occurs with double bonds-possibility that molecule will have different geometries; two different molecules with slightly different properties.
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3 years ago

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Can an ether form hydrogen bonds in the same way as does an ester

belka [17]

Yes, an OH group from ethanol can form a hydrogen bond to the ether O atom in the same way as it can do so with the single-bonded O atom in the ester.

The O atom in the carbonyl group of the ester can also form H-bonds with ethanol.

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3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

3 years ago

Can a solution with undissolved solute be supersaturated

Helen [10]

Answer: A supersaturated solution will not contain undissolved solute because the undissolved solute will be indicative of saturated solution.

Explanation:

A supersaturated solution is the one that consists of more than the maximum concentration of the solute in the solvent that is being dissolved at a given temperature. A saturated solution is the one in which the maximum concentration of solute has been dissolved in the solvent and no additional solute can be dissolved further.

According to the given statement, a solution with undissolved solute is a saturated solution rather a supersaturated solution.

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3 years ago