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3 years ago

A hypothesis is a statement that can be tested through a scientific investigation. What is the purpose of writing a hypothesis?

Chemistry

2 answers:

netineya [11]3 years ago

6 0

Answer: The purpose of a hypothesis is to find the answer to a question.

Explanation:

A hypothesis is a statement which is written as a source of prediction which is based upon the assumption of the cause of a natural event or process. The purpose of the hypothesis is basically to find the relevant answers to the scientific question. It is based upon few evidences and the entire statement of the hypothesis is tested using experimental procedures.

yaroslaw [1]3 years ago

6 0

The purpose of forming a hypothesis is to make an estimation as to what will happen during or the result of the scientific experiment!

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Use the following graph of a car traveling on a straight northerly path to answer this question. At what time would the

BARSIC [14]

Answer:

B 144.0 s is the best answer of this question

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4 years ago

How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Whitepunk [10]

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl = 1.4kg = 1400g

Unknown:

Number of ions of sodium = ?

Solution:

The compound NaCl in ionic form can be written as;

NaCl → Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

Now, let us find the number of moles in NaCl;

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

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles = $\frac{1400}{58.5}$ = 23.93mol

So;

Since 1 mole of NaCl gives 1 mole of Na⁺

In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance = 6.02 x 10²³ ions of a substance

23.93 mole of a substance = 6.02 x 10²³ x 23.93

= 1.44 x 10²⁵ ions of Na⁺

7 0

3 years ago

What is the mass of an object that has a density of 2.3 g/mL And a volume of 343ML

Bogdan [553]

Answer:

788.9g

Explanation:

density=m/v

2.3g/ml=m/343ml

=343×2.3

=788.9g

7 0

3 years ago

Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c

Margaret [11]

First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

∴ Pressure :

(A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g

∴ the rank of diffusion:

A > B > C

D) The rank of the Total kinetic energy of the molecules:

when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules

∵ Mw A < Mw B < Mw C

∴no .of molecules of A > B >C

∴ the rank of total kinetic energy is:

A > B > C

e) the rank of density:

when ρ = m/ v

and m is the mass & v is the volume and we have both is the same for A, B, and C

so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C

∴the rank of the collision frequency is:

A > B > C

6 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago