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Kaylis [27]
3 years ago
8

How many moles of carbon are in 3.5 L?

Chemistry
1 answer:
SashulF [63]3 years ago
8 0

Answer:

0.16mole

Explanation:

To solve this problem, we are going to assume that the number of moles of carbon to be determined is that at STP, standard temperature and pressure.

The number of moles of a substance at STP is given as;

 Number of moles  = \frac{volume}{22.4}  

Given volume  = 3.5L

Now, insert the parameters;

     Number of moles  = \frac{3.5}{22.4}   = 0.16mole

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It is true

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All of the alkali earth metals, Group 2, have two valence electrons. Which of these would represent the oxidation number of the
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Your answer should be C.) +2. "All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2." 


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5 0
3 years ago
For the reaction: 2 H2 + 2 NO → N2 + 2 H2O the observed rate expression, under some conditions, is: rate = k[H2][NO]2 Which of t
Andrew [12]

Answer:

Step 1 H2 + 2 NO → N2O + H2O (slow)

step 2 N2O + H2 → N2 + H2O (fast)

Explanation:

It is known that the slowest step in a reaction is the rate determining step in a sequence of reactions (reaction mechanism).

We have two important pieces of information in the question to guide our decision making process.

The overall reaction equation, and the rate expression. The two;

2 H2 + 2 NO → N2 + 2 H2O and rate = k[H2][NO]2 all support the answer given above.

8 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

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Hydroxide ion concentrstion in bleach is higher
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