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kkurt [141]
3 years ago
7

Is Tetracarbon decoxide ionic or covalent?

Chemistry
1 answer:
Cloud [144]3 years ago
8 0

Answer: covalent bond

Explanation: Tetracarbon Hexahyride is built using covalent bonds

Hope this helps please mark brainlist it would be greatly appreiciated :)

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Can someone give me a answer on why should renewable resources be used wisely?
irina [24]

We should use renewable resources wisely because <u>if we over use them the resources we already have will decline.</u>

5 0
2 years ago
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Q05 the "atomic number" of an atom is determined by the number of ________ it has.
Alexxx [7]
The atomic number of an atom is determined by the number of protons it has..

It is also the whole number shown on the periodic table 
6 0
3 years ago
You have an aqueous solution of chromium(III) nitrate that you titrate with an aqueous solution of sodium hydroxide. After a cer
barxatty [35]

Answer:

The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex, Cr(OH)4

Explanation:

The following reaction takes place when chromium(III) nitrate reacts with NaOH:

Cr(NO)_{3} +3 NaOH → Cr(OH)_{3} (s)+ NaNO_{3}

The precipitate that is formed is chromium hydroxide, Cr(OH)_{3}

When more NaOH is added, the precipitate reacts with it which then results in the formation of a soluble complex ion:

Cr(OH)_{3}(s) + OH^{-}(aq) → Cr(OH)_{4} ^{-}(aq)

Cr(OH)_{4} ^{-} is soluble complex ion

7 0
3 years ago
How much heat is released when you condense 93.9 g of water vapor? <br><br> Show work please ❤️
sammy [17]

<u>Answer:</u>

211.9 J

<u>Explanation:</u>

The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.

The amount of heat released can be represented by the formula:

Q=mL_e

where Q = heat energy, m = mass of water and L = latent heat of evaporation.

The latent heat of evaporation for water is L=2257 kJ/kg and the mass of the water is m=93.9 g=0.0939 kg.

The amount of heat released in this process is:

Q=mL_e = (0.0939kg)(2257 kJ/kg)= 211.9 J

7 0
3 years ago
S8 + 24 F2 ⟶ 8 SF6
Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

= 425g/256g/mol.      = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.  

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

   = 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆  to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)  

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.        

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.    

____________________

Here's another example just for grins ...

             C₂H₆O   +   3O₂     =>     2CO₂    + 3H₂O

Given:    253g          307g               ?               ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced        

Limiting Reactant

moles  C₂H₆O = 253g/46g/mol = 5.5 moles  => 5.5/1 = 5.5

moles  O₂ = 307g/32g/mol = 9.6 moles         =><em>  9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.  

 C₂H₆O   +       3O₂          =>     2CO₂    + 3H₂O

------------ 9.6 mole (L.R.)              ?               ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole  (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole  = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient  C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0
3 years ago
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