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4 years ago

What information does Jaime need to evaluate the accuracy of his calculation?

Chemistry

2 answers:

tiny-mole [99]4 years ago

8 0

Answer:

A. the accepted value for the density of aluminum

Explanation:

To know if your calculation is accurate or not, you need to know the accepted value. An example is if the accepted value was 15 and your calculation was 14.9, your calculation is accurate because its close to the actual value. If your calculation was 7, you would not be accurate at all because you are far from the actual value. Hope this helps please give brainliest!

Effectus [21]4 years ago

4 0

Answer:

well we can't awnser this we don't have the full information srry♥️✨

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BRAINLIST!!

Iteru [2.4K]

Answer:

as the flows along the outside edge of a curve it removes rock and soil from the river bank

6 0

3 years ago

A 118-ml flask is evacuated and found to have a mass of 97.129 g. when the flask is filled with 768 torr of helium gas at 35 ?c,

Inessa05 [86]

The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.

To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.

From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.

1) From pV = nRT, n = pV / RT

Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K

n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol

mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042

=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol

Now from a periodic table or a table you get that the molar mass of He is 4g/mol

So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.

7 0

4 years ago

A gas is at a pressure of 3.70 atm. What is this pressure in kilopascals?

Yanka [14]

Answer:
= 374.90 kPa

Calculation:
As we know atm and kiloPascal are related to each other as,

1 atm = 101.325 kPa
So,
3.70 atm = X
Solving for X,
X = (3.70 atm × 101.325 kPa) ÷ 1 atm

X = 374.90 kPa

7 0

4 years ago

How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

7 0

3 years ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

3 years ago