Which of the following patients should be admitted as an inpatient at a

Select the objects that hold the same amount of liquid as a 160−fluid-ounce jug. Select all that apply. A four 1−quart bottles B

Finger [1]

B, D and E

Explanation:

conversion factors

1c = 8oz

1pt = 2c

1qt = 2pt

For A and B

ounces to cup = 160/8 = 20c

cup to pints = 20c / 2c = 10pt

pint to quarts = 10pt/2pt = 5qt

B applies as four 1-quart and two 1-pt = 5 1-quart

Considering C

4 8oz = 32oz

160-32 = 128oz /8 = 16c/2 = 8pints C does not apply

considering D

8 * 8 = 64oz

160 - 64 = 96oz/8 = 12c/2 = 6pt/2 = 3qt

So D applies

E applies

5 0

2 years ago

Is there an N−Cl bond in solid ammonium chloride?

Tems11 [23]

When ammonia is reacted with HCl it abstracts proton from acid and forms Ammonium Ion and Chloride Ion.

NH₃ + HCl → ⁺NH₄ + Cl⁻ (simply Written NH₄Cl)
Structure,
The structure of Ammonium Chloride is among those structures which contains all three types of bonding's, i.e.

Ionic Bond

Covalent Bond

Coordinate Covalent Bond

Three Hydrogen atoms previously bonded with Nitrogen are covalent in nature. The new incoming proton from HCl forms co-ordinate covalent bond with Nitrogen and Chloride Ion containing negative charge make Ionic Bond with the positive Ammonium Ion. In question, if the line between Nitrogen and Chlorine atom is assumed covalent then it is incorrect. Structure is shown below,

5 0

2 years ago

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$