A molecule that oxidized gains electrons and energy. true or false

Use the periodic table to identify the element indicated by each electron configuration by typing in the

EleoNora [17]

Answer:

1s22s22p6: Neon (Ne)

1s22s22p63s23p3: Phosphorous (P)

1s22s22p63s23p64s1: Potassium (K)

1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)

1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4

Explanation:

you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number

5 0

2 years ago

Ok this is so weird but like how do you expose fake friends for talking about you? I know they did it but i dont know how to tal

murzikaleks [220]

Answer:

just do it

Explanation:

8 0

3 years ago

Read 2 more answers

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

Ionic bonds are formed by the attraction between two ________________.

cricket20 [7]

It would be C) atoms of the opposite charges. iconic bonds are formed by the attraction of two atoms of the opposite charge.

Hope this helps :)

can you please make this the brainiest answer it would really help me. thank you :)

7 0

3 years ago

Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

(0.75M) x (0.094L) = C2

(0.329L)

C2 = 0.214 M (Rounded)

When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.

7 0

2 years ago