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netineya [11]
3 years ago
12

In which block of the periodic table should iron be placed

Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0
Iron should be in the 3d block

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How many ions does each of these symbols have, I need answer and my science teacher doesn’t tell us or explain it to us
seropon [69]

Answer: about 14

Explanation:

6 0
3 years ago
Which discovery did J. J. Thomson make that improved upon Dalton's atomic theory?
erik [133]

Answer: Atoms contain tiny, negatively charged electrons

Explanation: Thomson's experiments with cathode ray tubes helped him to discover the electron (which Dalton did not know about). Dalton thought that atoms were indivisible particles, and Thomson's discovery of the electron proved the existence of subatomic particles.

3 0
3 years ago
How does the electron configuration of elements within the same group compare? (A.) They all have their valence electrons in the
lesya [120]

Answer:

B.They all have their valence electrons in the same type of subshell

Explanation:

  • The electron configurations of elements in the same group (column) of the periodic table have them in the same type of subshell.
  • But the subshells may be of different shells. Thus , the energies of them need not be the same.
  • For example , The Alkalai Metals are found in the first column of the periodic table Group IA. This set of elements all have valence electrons in only the 's' orbital and because they are in the first column they all have s^{1} configuration. i.e,
  1. H 1s^1
  2. Li 1s^2 2s^1
  3. Na 1s^2 2s^2 2p^6 3s^1
  4. K 1s^2 2s^2 2p^6 3s^2 3p^64s^1
5 0
3 years ago
The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 oC. What is the new temp
attashe74 [19]

Answer:

Explanation:

Given:

V1 = 200 ml

T1 = 20 °C

= 20 + 273

= 293 K

P1 = 3 atm

P2 = 2 atm

V2 = 400 ml

Using ideal gas equation,

P1 × V1/T1 = P2 × V2/T2

T2 = (2 × 400 × 293)/200 × 3

= 234400/600

= 390.67 K

= 390.67 - 273

= 117.67 °C

7 0
3 years ago
The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
svp [43]
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass 
                                      = 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water 
                                        = 4.7 / 0.88 = 5.34 moles 
∴ moles of the solution = total moles in solu - moles of water 
                                       = 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
                                                    = 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
                                           = 0.07 mol
moles of methanol = mass weight / molar mass 
                                 = 39.9  g / 32  g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
∴Pv(solu) = 84.55 mmHg


 
7 0
3 years ago
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