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4 years ago

Ethanol is an example of an ____________ (carboxyl/alcohol).

Chemistry

1 answer:

Leno4ka [110]4 years ago

5 0

it cannot be carboxyl..... ethanol cannot be an example of carboxyl

and ethanol is an alcohol ..... so therefore the answer is alcohol

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What element has an atomic mass 4?

Daniel [21]

Answer:

Helium has atomic mass 4.002602.

So the answer to yr question is helium.

4 0

3 years ago

Read 2 more answers

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

erica [24]

Answer: The amount of $P_4O_{10}$ formed is 469.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

8 0

3 years ago

How would you ask a girl out. I know that I can use my voice, but I don’t know what to say.

kirill [66]

Answer:

Just go out there and be yourself brother!

Explanation:

If she doesn't like you, she doesn't like the real you, therefore you don't deserve her! lol

4 0

3 years ago

The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=

castortr0y [4]

Answer:

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

Step 1: Data given

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=−907kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH=−113kJ 3NO2+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−139kJ

Step 2: Multiply equations

Multiply the first equation by 3:

12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(l) ΔH = −2721 kJ

Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g) ΔH = −556 kJ

Step 3: Get the equations together

12 NH3(g) + 15 O2(g) + 12 NO(g) + 6 O2(g) + 12 NO2(g) + 4 H2O(l) →

12 NO(g) + 18 H2O(l) + 12 NO2(g) + 8 HNO3(aq) + 4 NO(g)

ΔH = −2721 kJ − 678 kJ − 556 kJ

We can simplify as followed:

12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

Step 4: Determine the total energy change for the production of one mole of aqueous nitric acid by this process:

−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

3 0

3 years ago

(110.3 g) + (45.27 g) + (54.43 g)

Darina [25.2K]

Answer:

210 grams

Explanation:

just add them all up

7 0

3 years ago