All categories

3 years ago

How can a chemical reaction be sped up?Select from the drop-down menu to correctly complete the statement.

Chemistry

2 answers:

viktelen [127]3 years ago

8 0

Answer: Temperature

Explanation:

RoseWind [281]3 years ago

6 0

Answer:

Concetration

Explanation:

You might be interested in

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

Identify the false statement from the following.

LUCKY_DIMON [66]

The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.

<h3>What are the factors that affect London dispersion forces?</h3>

Generally, the factors which affects the London dispersion forces a dispersion force are as follows:

Shape of the molecules
Distance between molecules
Polarizability of the molecules

However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.

So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement

Learn more about London dispersion forces:

brainly.com/question/1454795

7 0

2 years ago

Please help due in 5 min.

Lina20 [59]

Answer:

B) Hope this helps! :)

Explanation:

6 0

3 years ago

A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show t

Oxana [17]

This question is incomplete, the complete question;

A reaction between liquid reactants takes place at 16°C in a sealed, evacuated vessel with a measured volume of 10.0 L . Measurements show that the reaction produced 37.g of dinitrogen monoxide gas.

Calculate the pressure of dinitrogen monoxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.

Answer:

Explanation:

Given that;

Volume = 10.0 L

Temperature T = 16°C = ( 16 + 273) = 289 K

we know that molar mass of N₂O = 44 g/mol

number number of moles n will be: weight/molar mass = 37/44 = 0.8409 moles

Universal Gas constant R = 0.0821 atm.mol⁻¹.k⁻¹

pressure p = ?

from the ideal gas law

PV = nRT

P = nRT / n

so we substitute

P = (0.8409 mole × 0.0821 atm.mol⁻¹.k⁻¹ × 289 K) / 10.0 l

P = 19.95195 / 10.0

P = 1.9952 atm

6 0

3 years ago

Help me please!<br> ASAAAAAAAAAAAAAAAAAAAP

Anna [14]

can you zoom in a bit more

4 0

3 years ago