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mrs_skeptik [129]
3 years ago
10

Which formula represents a hydrocarbon with a double covalent bond?

Chemistry
1 answer:
Nataliya [291]3 years ago
4 0

Answer:

c is the answer

step by step explanation is not

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If 30.943 g of a liquid occupy a space of 35.0 ml. What is the density of the liquid in g/cm3?
timofeeve [1]
Mass / volume = density
30.943g / 35ml = 0.88408571g/ml 

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Answer:

It should be D.

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3. How can a community help implement Ecological Solid Waste Management.
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It help cause we compost food scraps and other organic wastes. We also reuse and recycle materials to organize for government and industry to develop community recycling materials.

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2 years ago
How much heat does it take to melt 5.0g solid copper?
Leya [2.2K]

Answer:

Q=1.04kJ

Explanation:

Hello,

In this case, for latent heat (phase change) we need to consider the enthalpy associated with the involved process, here, melting or fusion; thus, the enthalpy of fusion of copper is 13.2 kJ/mol, therefore, the heat is computed as:

Q=m\Delta H_{fus}

Nevertheless, since the given enthalpy is per mole of copper, we need to use its atomic mass to perform the correct calculation as follows:

Q=5.0g*\frac{1mol}{63.54}* 13.2\frac{kJ}{mol}\\ \\Q=1.04kJ

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3 years ago
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t
Sloan [31]

Answer:

Mass =  42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2  = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass =  42.8g

6 0
3 years ago
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