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2 years ago

Why does increasing the number of trials increase confidence in the results of the experiment?

2 answers:

5 0

It rises confidence for the reason that the more times you conduct the similar experiment over and over should either demonstrate your hypothesis right and wrong and remove any random incidences that might touch your results. Meaning it permits to have a more accurate measure or conclusion.

Pavel [41]2 years ago

4 0

The answer is...

Sample Response: Increasing the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.

I just took the test.

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Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is

Citrus2011 [14]

Depression in freezing point (Δ $T_{f}$ ) = $K_{f}$ ×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$ ,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$ )

Thus, (Δ $T_{f}$ ) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ $T_{f}$ ) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ $T_{f}$ ) = - $0.03162^{0}C$

8 0

2 years ago

Read 2 more answers

A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle

marin [14]

Answer:

=37.83783784

Explanation:

Find the total sum of all coins,

which is 37, take the number of pennies and the total of all coins put in parenthesis( 14/37) like so and than * times them by 100

you equation should look like this

(14/37)* 100= and than the answer shown above should be the one you received. I have checked this with multiple calculators, it should be accurate.

8 0

2 years ago

Read 2 more answers

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C

UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M 0.24 M 0.48 M 0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0

2 years ago

Express the quantity 556.2 x 10^-12 in each units A.) ms B.) ns C.) ps D.) fs

zavuch27 [327]

The answer:
we should know the meaning of each abbreviation:
ms means millisecond, its value is 10^-3 s
ns means means nanosecond, its value is 10^-9 s
ps means picosecond, its value is 10^-12 s
fs means femtosecond, its value is 1x 10^15 s

Expressions of the quantity 556.2 x 10^-12 are
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
556.2 x 10^-12 = 556.2 x 10^- 27 x 10^15 = 556.2 x 10^- 27 fs

6 0

2 years ago

The volume of hcl gas required to react with excess ca to produce 11.4 l of hydrogen gas at 1.62 atm and 62.0 °c is ________ l.

seropon [69]

Answer:

22.8 L

Step-by-step explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem:

Gases at the same temperature and pressure react in simple whole-number ratios.

1. Write the chemical equation.

Ratio: 2 L 1 L

Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)

V/L: 11.4

2. Calculate the volume of HCl.

According to the law, 2 L of HCl form 1 L of H₂.

Then, the conversion factor is (2 L HCl/1 L H₂).

Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)

= 22.8 L HCl

3 0

2 years ago