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icang [17]
3 years ago
10

Help plz and will give brainly

Chemistry
1 answer:
Rashid [163]3 years ago
3 0
I think D or C is my top answers
You might be interested in
Covalent bonding occurs between two oxygen atoms to form O2. What's the partial charge on each oxygen atom in 02?
Alik [6]

Answer:

Both oxygen atoms have zero partial charge.

Explanation:

In covalent bonds, electrons are shared between the two bonding atoms. The shared electron pair of the covalent bond is positioned between the nuclei of the two bonding atoms.

A covalent bond may be formed between two or more elements of different electronegativities. When a difference in electro negativity exists between these atoms in a covalent bond, the molecule becomes polarized. The shared electron pair of the covalent bond becomes closer to the nucleus of the more electronegative atom (s). This more electronegative atom (s) becomes partially negative while the other atom becomes partially positive.

When the two bonding atoms are of equal electronegativities such that the electro negativity difference between the bonding atoms is zero, there is now no difference in electro negativity and no consequent partial charges.

Since the two oxygen atoms have the same electro negativity, there is no difference in electronegativity, hence there is no partial charge between the two oxygen atoms.

3 0
3 years ago
If 40.0 g of molten iron(II) oxide reacts with 10.0 g of mag-nesium, what is the mass of iron produced
vagabundo [1.1K]

Answer:

m_{Fe}=23.0gFe

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

FeO+Mg\rightarrow Fe+MgO

Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:

n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO

Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:

n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO

Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:

m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\  \\m_{Fe}=23.0gFe

Regards.

8 0
4 years ago
A rigid container of O 2 has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273°C ?
Vinvika [58]

Answer:

260 kPa

Explanation:

To answer this problem we can use <em>Gay-Lussac's law</em>, which states that at a constant volume (such as is the case with a rigid container):

  • P₁T₂=P₂T₁

Where in this case:

  • P₁ = 340 kPa
  • T₂ = 273 °C ⇒ 273 + 273.16 = 546 K
  • P₂ = ?
  • T₁ = 713 K

We <u>input the data</u>:

  • 340 kPa * 546 K = P₂ * 713 K

And <u>solve for P₂</u>:

  • P₂ =  260 kPa
5 0
3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
If D+2 would react with E-1, what do you predict the formula to be. A) DE B)D2E C)DE2 D) D2E2
allochka39001 [22]
All you need to do the cross the charges and put it as suffixes. 

D+2 + E-1 ----> DE2

answer is C
5 0
3 years ago
Read 2 more answers
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