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Tpy6a [65]
3 years ago
10

The more bonds an atom can make, the more likely it is to combine with other atoms in different ways Which element is most likel

y able to make the greatest variety of bonds? a. nitrogen b. hydrogen c. oxygen d. carbon​
Chemistry
2 answers:
avanturin [10]3 years ago
8 0

Answer:

b                                 xplanation:

Semenov [28]3 years ago
5 0

Answer:

D. Carbon​.

Explanation:

Carbon can be mixed with Many things. for example, YOU are mostly made out of carbon. Other things include furniture, clothing, plastics, charcoal, pencil led, diamonds, and gas.

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154 pm is greater than 7.7 x10^-9<br> A. True<br> B. False
Nesterboy [21]
The answer is false.
7 0
3 years ago
A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

5 0
3 years ago
In CaSO4, the oxidation number of Ca is<br> that of S is<br> and that of O is
Assoli18 [71]

Answer: In CaF2, the oxidation number of Ca is +2

, and that of F is -1

. In H2SO4, the oxidation number of H is +1

, that of S is +6

, and that of O is -2

. In CaSO4, the oxidation number of Ca is +2

, that of S is +6

, and that of O is -2

. In HF, the oxidation number of H is +1

, and that of F is -1

7 0
3 years ago
2 questions<br> Popeyes or KFC<br> Mcdonalds, Wendys, Burger King, Shake Shack, or Chick Fil A
Ymorist [56]

Answer:

Chick-fli-a

Explanation:

3 0
3 years ago
Read 2 more answers
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
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