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4 years ago

Relationship between boiling temperature and vapour pressure

Chemistry

1 answer:

Ronch [10]4 years ago

3 0

Answer:

The higher the vapor pressure of a liquid at a given temperature, the lower the normal boiling point of the liquid.

Explanation:

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Question: Which is the last step in excavation of the skeletal remains?

Kay [80]

Answer: B:

Explanation:

This is the most reasonable answer

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3 years ago

Read 2 more answers

Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO

stiv31 [10]

Answer:

NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given data:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂ → 4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃ : NO NH₃ : H₂O

4 : 4 4 : 6

2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂ : NO O₂ : H₂O

5 : 4 5 : 6

4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide = number of moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

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3 years ago

Identify the missing reagents 1 & 2; draw the final organic product(s)<br> c.

abruzzese [7]

Co2 is correct buddy

3 0

4 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$