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romanna [79]
3 years ago
9

Which of the following changes would increase the rate of a reaction between vinegar and sodium hydroxide? (2 points) Question 2

1 options: 1) heating up the vinegar 2) diluting the vinegar 3) cooling the sodium hydroxide 4) decreasing the volume of the container Save
Chemistry
2 answers:
posledela3 years ago
8 0
I would do option 1) heat up the vinegar. It would increase the rate of reaction.

aleksandr82 [10.1K]3 years ago
6 0

<u>Answer: </u>The correct answer is Option 1.

<u>Explanation:</u>

Rate of reaction can be increased by the following factors:

1. Concentration of the reactants: Greater the concentration of the reactants, faster is the reaction. Conversely, as the concentration of the reactants decrease, the rate of reaction also decreases.

2. Temperature: The rate of the reaction increases with increase in temperature.

3. Surface area of the reactants: For the reaction involving a solid reactant, the smaller is the particle size that is the greater is the surface area, faster will be the reaction.

From the given options, Temperature is playing the role in increasing the rate of the reaction.

Heating of vinegar will lead to the increase in the rate of the reaction because temperature is also increasing.

Hence, the correct answer is Option 1.

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Calculate the density of a sample of 1.00 mole of nh3 at 793mmhg and -9.00 c
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0.164 g/L is the density of a sample of 1.00 mole of NH_3 at 793mmhg and -9.00 degrees celcius.

<h3>What is density?</h3>

Density is the mass of a unit volume of a material substance. The formula for density is d = \frac{M}{V}, where d is density, M is mass, and V is volume.

Given data:

n = 1.00 mole

P=793 mm hg =1.04342 atm

T=-9.00 degree celcius = -9.00 + 273= 264 K

V=?

Using Ideal Gas Law equation:  

PV = n R T      

R = gas constant = 0.082057 L-atm/(mol-K)

(1.04342 atm)(V) = 5 X 0.082057 L-atm/(mol-K)  X 264 K

V = 103.67 Liters

Now calculate density:

Mole weight of NH_3 = 1.00 mole

So, the mass of NH_3 = 17.031 g

Density =  \frac{mass}{volume}  

Density =  \frac{17.031 g}{ 103.67 Liters}  

= 0.164 g/L

Hence, 0.164 g/L is the density of a sample of 1.00 mole of NH_3 at 793mmhg and -9.00 degrees celcius.

Learn more about the density here:

brainly.com/question/15164682

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Answer:

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