A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. Their percent error is 22.1%.
Percent error is a measure of the difference between an observed value and a true value.
Actual Speed (True Value) = 2.2 m/s
Experimental Speed (Calculated Value) = Distance / Time = 6 m / 3.5 s = 1.714 m/s
The formula for calculating percent error is:
Percent Error = ((|Actual Value - Experimental Value|) / |Actual Value|) * 100%
Calculate the absolute difference between the actual speed and the experimental speed:
|2.2 - 1.714| = 0.486
Calculate the absolute value of the actual speed:
|2.2| = 2.2
Percent Error = (0.486 / 2.2) * 100%
= 0.221 * 100%
= 22.1%
The calculated percent error is approximately 22.1%. This means that the lab group's calculated speed of 1.714 m/s is about 22.1% lower than the true speed of 2.2 m/s.
Percent error is a way to quantify the accuracy of experimental measurements. A positive percent error indicates that the experimental value is higher than the true value, while a negative percent error indicates that the experimental value is lower. In this case, since the calculated speed is lower than the true speed, we have a positive percent error.
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Answer:456
Explanation:
Answer : The volume of calcium hydroxide is, 32.89 ml
Explanation :
Using neutralization law,
where,
= basicity of an acid = 1
= acidity of a base = 2
= concentration of hydrobromic acid = 0.389 M
= concentration of calcium hydroxide = 0.0887 M
= volume of hydrobromic acid = 15 ml
= volume of calcium hydroxide = ?
Now put all the given values in the above law, we get the volume of calcium hydroxide.
Therefore, the volume of calcium hydroxide is, 32.89 ml
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution
The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.
To calculate the molarity (M) of a solution, you can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
a. 0.38 moles of LiNO₃ in 6.14 L of solution:
Molarity (M) = 0.38 moles / 6.14 L = 0.062 M
b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:
First, you need to convert grams to moles using the molar mass of C₂H₆O.
Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol
Now, calculate moles of C₂H₆O:
moles = 72.8 g / 46.08 g/mol = 1.58 moles
Molarity (M) = 1.58 moles / 2.34 L = 0.675 M
c. 12.87 mg of KI in 112.4 mL of solution:
First, convert milligrams to grams (1 g = 1000 mg):
12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)
Now, convert mL to liters (1 L = 1000 mL):
112.4 mL = 0.1124 L
Calculate moles of KI:
Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol
moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles
Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M
So, the molarities of the solutions are as follows:
a. 0.062 M
b. 0.675 M
c. 0.000688 M
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Answer:
So trees in temperate don't lose their leaves because the weather events aren't harsh enough.
Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.
Explanation:
Ok so first we'll define some things
Deciduous Trees= Trees that lose all of their leaves for part of the year.
Trees shed their leaves trees to try and survive harsh weather events.
Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.
Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.
Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.
Answer:
Hi,
The assembling of proteins starts with the attachment of m RNA to the ribosomes in the cytoplasm. Each of the ribosome read the code in the m RNA from “start” to “ stop” choosing the specific amino acid building block and removing the unwanted growing protein. The ribosome performs this process in 0.02 seconds and with this rate it is possible for the cell to perform assembling of small protein such as insulin.
Best wishes!
Answer:
A, D, E, F
Explanation:
just took it on edge
The expression for the equilibrium constant Kp for the following reaction is
It refers to the ratio of the concentration of products to the concentration of reactants where each raised to the power of their stoichiometric ratios. It is expressed as K.
Since the given expression is
MoO3(s) ↔ 2 MoO2(s) + O2(g)
So, the above expression should be considered for constant kp
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Answer:
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
is the constant of a certain reaction at equilibrium for gaseous reactants and products.
For the given chemical reaction:
The expression of for above equation follows:
As solids do not exert pressure, and are not involved.
Answer:
it’s electron configuration is 1s^2 2s^2 2p^4. To determine valence electrons, add the outermost s and p orbitals. In an oxygen atom, 8 electrons are present. Electron present in the first shell (n=1) 2n^2=2 (1)^2=2 (1)=2.