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stich3 [128]
3 years ago
10

A student calculates the mass of a piece of copper to be 8.3 grams. The actual mass of the copper is 10. grams. What is the perc

ent error? Show your work below. Round your answer to two digits
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
3 0

<u>Answer:</u>

17%

<u>Explanation:</u>

Mass of a piece of copper calculated by student = 8.3 grams

Actual mass of copper = 10 grams

We know the formula for the percent error:

Percentage error = (| experimental value - actual value | / actual value) * 100

So substituting the given values in the above formula:

Percent error = (|8.3 - 10| / 10) * 100 = (1.7/10) * 100 = 17%

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A sample of methane gas, CH4, occupies 3.25 L at temperature of 19.0 o C. If the pressure is held constant, what will be the tem
Artemon [7]

Answer:

625.46 °C

Explanation:

We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 19 °C

T(K) = 19 °C + 273

T(K) = 292 K

Next, we shall determine the Final temperature. This can be obtained as follow:

Initial volume (V₁) = 3.25 L

Initial temperature (T₁) = 292 K

Final volume (V₂) = 10 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

3.25 / 292 = 10 / T₂

Cross multiply

3.25 × T₂ = 292 × 10

3.25 × T₂ = 2920

Divide both side by 3.25

T₂ = 2920 / 3.25

T₂ = 898.46 K

Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 898.46 K

T(°C) = 898.46 – 273

T(°C) = 625.46 °C

Therefore the final temperature of the gas is 625.46 °C

4 0
3 years ago
Rank the following elements in order of decreasing atomic radius (1 is biggest, 5 is smallest)
AlexFokin [52]

There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

<h3>What is periodic table?</h3>

Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.

Along the period, the size of elements decreases. Down the group the size of elements increases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

Therefore, atomic radius in decreasing order is Bi>Sb>As>N>O.

Learn more about periodic table, here:

brainly.com/question/11155928

#SPJ1

3 0
1 year ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
...............………...mkmlkml;m
bagirrra123 [75]

Answer:

390

Explanation:

Specific heat capacity= heat/mass × temperature

x = 2925 \div 0.5 \times 15 = 390

Remember you convert gram into kilogram and 1 gram =0.001 kilogram

5 0
2 years ago
How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f
BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

A=\text{ 50 }\times2^{\frac{-t}{14.3}}

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}

So, after 84 days the P-32 remaining will be 0.85 mg

4 0
1 year ago
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