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stellarik [79]
3 years ago
10

Methane is combusted with Oxygen to produce Carbon Dioxide and Water. How many grams of oxygen are required to react with 55 gra

ms of
Methane?
A.55 g
B.110 g
C.165 g
D.220 g

Chemistry
1 answer:
Rom4ik [11]3 years ago
7 0
The answer is B.110 g
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Why does hydrochloric acid have a higher boiling point than diatomic fluorine? The stabilizing effect of dipole interactions in
il63 [147K]
The correct answer would be the fourth option. Hydrochloric acid has a higher boiling point than fluorine molecule because the former is a polar molecule which means that it has stronger bonds than fluorine which is a non-polar molecule. Stronger bonds require more energy to break these bonds.
6 0
3 years ago
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URGENTTTTT What is the correct formula when Al+3 combines with CrO4-2
PIT_PIT [208]
A and C are incorrect because they are not complete transfer of valence electrons. Ionic bonds best to form a neutral molecule
8 0
3 years ago
Will the ph of a solution of nh4cn be >7, <7, or =7?
never [62]
PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid. 
NH₄CN + H₂O ⇒ NH₃ + HCN 

NH₄⁺ + H₂O -----> H₃O⁺ + NH₃

CN⁻ + H₂O -----> HCN + OH⁻ 

Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:

Ka(HCN) = 4.9 x × 10⁻¹⁰

Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴

Let's first calculate Ka for NH₄⁺: 
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw

</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰

Then, Kb for CN⁻:

Kb(CN⁻) x Ka(HCN) = pKw

Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵

 
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.


7 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
How many moles are there in 24.00 g of NaCl
Elis [28]

Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

               58.5 g of NaCl ------------------- 1 mol

               24.0 g               ------------------- x

                     x = (24 x 1) / 58.5

                     x = 0.41 moles

6 0
3 years ago
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