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3 years ago

I need help with question 5 whoever helps could get Brainlest or extra credit

Chemistry

1 answer:

Scrat [10]3 years ago

3 0

1 is c 2 is b 3 is 1 d 4 is a

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The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The stand

ale4655 [162]

Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

7 0

3 years ago

Copper produces a green flame test. When is the green light emitted?

belka [17]

Answer: When you burned the skewer tip coated with copper sulfate the green light is emitted

Explanation:

when the copper is burned it makes a green light

8 0

3 years ago

Determine the electron configuration of nitrogen numerically.

Vera_Pavlovna [14]

Answer: 1s2 2s2 2p3

Explanation:

atomic number is 7 hence it shuld be 2,5 if numerically.. tho the standard notation is the above giver answer

5 0

3 years ago

Read 2 more answers

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

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3 years ago

To be effective, a certain drug must not drop below 15 mg in the body. A 120

frez [133]

Thus, it follows that after 4 to 5 half-lives, the plasma concentrations of a given drug will be below a clinically relevant concentration and thus will be considered eliminated. Conversely, the accumulation of a drug can reach a steady-state during an infusion

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3 years ago