Ask question

Ask question

All categories

3 years ago

14

Identify two ways that the rate of a chemical reaction could be increased

1 answer:

Svetlanka [38]3 years ago

7 0

Increase the Surface Area of the Reactants. Increasing the surface area of the reactants increases the rate of the reaction. More surface area means more collisions of the reactant molecules and an increased rate of the reaction. This occurs when reactants are made to react in powdered form

You might be interested in

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

Grace [21]

Answer:

0.21 g

Explanation:

The equation of the reaction is;

NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)

Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles

Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles

Since the reaction is 1:1, NaCl is the limiting reactant.

1 mole of NaCl yields 1 mole of AgCl

0.00147 moles of NaCl yields 0.00147 moles of AgCl

Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol

= 0.21 g

3 0

2 years ago

Potassium-40 (40k) is a radioactive isotope of potassium (atomic number 19). how many neutrons are located in the nucleus of pot

sineoko [7]

Number of neutrons = mass number - atomic number = 40 - 19 = 21 answer

6 0

3 years ago

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

2 years ago

A hypothetical element, E, has two stable isotopes. One isotope has a natural abundance of 68.037% and has an atomic mass of 46.

I am Lyosha [343]

If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.

Substituting into the atomic mass formula,

$47.574=(46.449)(0.68037)+x(1-0.68037)\\\\ 47.574=31.60250613+0.31963x\\\\15.97149387=0.31963x\\\\x \approx \boxed{49.969 \text{ u}}$

6 0

2 years ago

A sample of a substance has a mass of 7.5 g and a volume of 2.5 cubic centimeters. What is the object’s density?

Inga [223]

Answer:

option A is the correct answer .

Explanation:

as density = mass per unit volume

density = 7.5/2.5 = 3 gm / cm³ ..is the answer ...

pls mark my answer as brainlist plzzzz and plz vote

7 0

2 years ago

Read 2 more answers