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2 years ago

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A researcher recorder that a chemical reaction require 2.90 gal water and raised the temperature of the reaction vessel by 57°F.

Convert the researchers volume measurement to the SI unit of liters.
There are 3.75 L in 1 gallon

A. 8.9 L
B. 9.11 L
C. 11 L
D. 11.0 L

1 answer:

Ivan2 years ago

4 0

Answer:

10.875L

Explanation:

The problem here is a simple conversion. The conversion is from gal to liters Liter is a SI unit for recording volume as stated in the problem.

Given that:

1 gallon of water = 3.75L

2.90 gallon will be 2.9 x 3.75; 10.875L

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PLEASE HELP ME!!!!!!!!

Mazyrski [523]

I think the answer is c

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2 years ago

Read 2 more answers

The air inside a bicycle tire pump has 27 joules of heat conducted away from while 77.9 joules of work done are being done on it

tatyana61 [14]

Answer:

51 J

Explanation:

The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.

77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J

We can calculate the change in the internal energy (ΔU) using the following expression.

ΔU = Q + W

ΔU = (-27 J) + 77.9 J

ΔU = 51 J

7 0

2 years ago

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

2 years ago

If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?

bija089 [108]

<h3>Molar mass:-</h3>

$\\ \sf\longmapsto HC_2H_3O_2$

$\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)$

$\\ \sf\longmapsto 1u+24u+3u+48u$

$\\ \sf\longmapsto 28u+48u$

$\\ \sf\longmapsto 76u$

$\\ \sf\longmapsto 76g/mol$

No of moles=0.2mol
Given mass=?

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}$

$\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}$

$\\ \sf\longmapsto Given\:Mass=0.2\times 76$

$\\ \sf\longmapsto Given\:Mass=1.52g$

7 0

2 years ago

A tank in the shape of a rectangular prism measures 10 dm by 4 dm by 6 dm. the tank is completely filled with 8908.8 l of liquid

nevsk [136]

1) The question contains an unknown unit

The number 8908.8 has to be in units of mass: for example, kg or grams.

Here you indicated L.

I am going to work assuming that L is a mass unit. So you can see the way to solve the problem, but you have to verifiy the real unit of the statement and substitute with it.

With that in mind you can find the density of the liquid from:

density = mass / volume

2) Calculate the volume.

The volume of the liquid is the volume of the vessel, because it is filled.

The volume of the vessel is calculated from the formula of volume for a rectantular prism.

Volume of a rectangular prism = area of the base * height = side * side * height

=> Volume = 10 dm * 4 dm * 6 dm = 240 dm^3 = 240 liter

3) Calculate the density:

density = mass /volume = 8,908.8 L / 240 liter = 37.12 L / dm^3

Answer: 37.12 L / dm^3

6 0

2 years ago