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3 years ago

A theory which repeatedly fails to confirm the expected predictions:

Chemistry

2 answers:

Leona [35]3 years ago

4 0

Answer:

Probably should be discarded.

Explanation:

Studentka2010 [4]3 years ago

3 0

A theory which repeatedly fails to confirm the expected predictions probably should be discarded. If the given theory is failing once or twice, then the facts and figures can be researched much more to understand the real cause of the error caused. But in case, the theory is again and again fails to confirms the expected predictions, then there is a gap in the facts and figures and the theory being studied and hence it should be discarded.

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A chemist places 2.5316 g of Na 2SO 4 in a 100 mL volumetric flask and adds water to the mark. She then pipets 15 mL of the resu

Lera25 [3.4K]

Answer:

The concentration of the most dilute solution is 0.016M.

Explanation:

First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:

$[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M$

First dilution

We can use the dilution rule:

C₁ x V₁ = C₂ x V₂

where

Ci are the concentrations

Vi are the volumes

1 and 2 refer to initial and final state, respectively.

In the first dilution,

C₁ = 0.178 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.178M \times 15mL}{50mL} =0.053M$

Second dilution

C₁ = 0.053 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.053M \times 15mL}{50mL} =0.016M$

3 0

3 years ago

Hey whats your fave Harry Potter character? Mines DEF Draco Malfoy!!!!!!

posledela

Answer:

Duh Draco lol

Explanation:

hes cute that's why lol

6 0

3 years ago

Read 2 more answers

QUESTION 3

Blababa [14]

These questions are all about indirect and direct variation with PV=nRT, the ideal gas equation

Q3.
false, because of PV=nRT, the ideal gas equation if V goes up, P has to go down to equal nRT

Q4. false, if V remains constant, and P and T are constant as moles of gas are added, then something is wrong becse something has to change when stuff is added (V has to go down)

Q5.
PV=nRT
when T and n are constant, (R is the gas constant)
PV=k, aka V=k/P which means inversly proportional

TRUE

Q6.
ggeasy
refer to past question
PV=k
if P is doubled then V has to halve in order to equal k
1/2 times 2=1
volume is halved

Q7. use charles law
V/T=k
so
given
V=4
T= kelvins, so 299
4/299=k
so when temp goes to 22 does V go to 3.95
4/299=3.95/295?
true
because they're equal

Q8
FALSE, must be used in kelvins
T=absolute tempurature in kelvins

Q9
PV=nRT
solve for T
(PV)/(nR)=T

use final volumes and pressures
P=5atm
V=24L
n=1
R=0.082057 atm L/(mol K)
(5atm*24L)/(1mol*0.082057 atm L/mol K)=T
see, if you didn't mess up, the units cancel nicely
T=1462.4
1200 K is closest

Q10
PV/T=constant because moles are constant (supposedly)

V=4L
P=2.08atm
T=275K

so find initial to final is constant
(2.08atm*4L)/(275K)=(Pfinal*2.5L)/(323K)
solve for Pfinal
Pfinal=3.92315 atm
answer is 3.9atm

Merry Christmas

4 0

4 years ago

Please help don’t understand this I attached the question

Rasek [7]

Answer:

58.32

Explanation:

The mass number is one mole and for Mg it is 24.306 for O it is 15.999 and for H it is 1.008. because O and H have 2, you multiply them by 2 and add all the numbers and it equals 58.32

8 0

1 year ago

Read 2 more answers

A 14.570 g sample of CaCl2 was added to 12.285 g of K2CO3 and mixed in water. A 3.494 g yield of CaCO3 was obtained.

natta225 [31]

Answer:

Limiting reagent is the potassium carbonate.

Percent yield of calcium carbonate is: 39.3 %

Explanation:

The reaction is:

CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl

Formula for percent yield is:

(Produced yield / Thoeretical yield) . 100

Firstly we determine the moles of each reactant, in order to say what is the limiting reagent: ratio is 1:1.

1 mol of chloride need 1 mol of carbonate.

14.570 g . 1 mol /110.98 g = 0.131 moles of CaCl₂

12.285 g . 1 mol / 138.2g = 0.0889 moles of carbonate.

Limiting reagent is carbonate. For 0.131 moles of CaCl₂ we need the same amount of carbonate and we have less moles.

Ratio is also 1:1, with calcium carbonate.

1 mol of potassium carbonate produces 1 mol of calcium carbonate

then, 0.0889 moles will produce the same amount of CaCO₃

We convert moles to mass: 0.0889 mol . 100.08g /mol = 8.89 g

That's the theoretical yield; to find the percent yield:

(3.494 g / 8.89g) . 100 = 39.3%

7 0

3 years ago