Help!!!!!! i have not done science since December!!!!!!! _________ is the distance traveled during a specific unit of time.

What’s the answer to this science problem?

Law Incorporation [45]

Answer:

The image is not accurate because

the heavier atoms are moving faster than the lighter ones

but the lighter ones must move faster than the heavier ones.

Explanation:

The image depicts the atoms of two different elements; atoms of lighter and heavier elements.

And the rate of diffusion (movement of atoms) of an element is inversely proportional to the square root of its molecular.

⇒ As the weight increases, the rate of diffusion (movement of atoms) decreases.

And as the weight decreases, the rate of diffusion (movement of atoms) increases.

so, heavier atoms move slowly compared to lighter atoms.

The image is not accurate because

the heavier atoms are moving faster than the lighter ones

but the lighter ones must move faster than the heavier ones.

5 0

3 years ago

What is the difference between repetition and replication of an investigation

schepotkina [342]

Repetition is repeating the same investigation over and over again . Such as you doing an experiment and redoing to to see if you get the same results however it is often confused with replication . Replication is when you copy an experiment done by someone else . Such as seeing an experiment and trying to get the same results as someone . Repetition you often do yourself while replication you copy someone else .

4 0

3 years ago

A sample of rest of the following composition 51.85 g of iron 22.22 g of oxygen 16.67 g of water

valentinak56 [21]

Answer:

To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:

%

C

9

mol C

×

molar mass C

molar mass

C

9

H

18

O

4

×

100

=

9

×

12.01

g/mol

180.159

g/mol

×

100

=

108.09

g/mol

180.159

g/mol

×

100

%

C

60.00

%

C

3 0

2 years ago

g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i

kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is $\alpha_{Y^{-4}}$ =0.81

So the Conditional Formation constant= $K_f$ =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL

V e= 25.00 mL

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.

$[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M$

${K^'}_f$

Ca²⁺ + Y⁴ ⇄ CaY²⁻

Initial 0 0 0.001667

change +x +x -x

equilibrium x x 0.001667 - x

${K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\$

$x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14$

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M

pca = —log(2 15x101= 6.7

3 0

3 years ago

When 1 mol of methane is burned at constant pressure, −890 kJ/mol of energy is released as heat. If a 3.72 g sample of methane i

Nonamiya [84]

Answer:

-206.9kJ

Explanation:

Heat of combustion = -890kJ/mol

Mass of methane burnt = 3.72g

Unknown:

Heat liberated = ?

Solution:

Heat of combustion is the heat liberated when one mole of an element or compound is burnt completely in excess oxygen gas.

To solve this problem;

Find the number of moles of methane;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CH₄ = 12 + 4(1) = 16g/mol

Number of moles = $\frac{3.72}{16}$ = 0.23moles

Now, heat liberated by 0.23mol of methane = -890kJ/mol x 0.23mol

= -206.9kJ

8 0

3 years ago