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mezya [45]
3 years ago
12

A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition o

f dilute HCl, no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown? The list to choose from: Cr(NO3)3 Co(NO3)2 NaNO3 Ni(NO3)2 Bi(NO3)3 Ca(NO3)2 Sn(NO3)4 Cd(NO3)2
Chemistry
1 answer:
finlep [7]3 years ago
5 0

Answer:

Explanation:

   The nitrates of Bi,Sn and Cd is ruled out because their sulfides are insoluble in acidic medium.

Nitrates of Ni or Co may be present because their sulfides are insoluble in basic medium. The presence of other nitrates are ruled out.

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Alice is fed through a gastric tube. She has developed a pressure wound and needs extra protein to support wound healing. The re
kumpel [21]

Answer:

modular

Explanation:

<em>The addition of extra liquid protein supplement to the tube feeding formula of Alice is an example of modular formula.</em>

<u>A modular formula consists of a deficient food supplements that is specifically rich in a particular macronutrient such as protein, carbohydrate, etc. </u>

The supplements in modular formula are usually in liquid from and are introduced into the tube feeding formula of a patient undergoing tube feeding.

In this case, the macronutrient in the liquid food supplement being given to Alice is protein.

3 0
3 years ago
Read 2 more answers
The decomposition reaction of carbon disulfide to carbon monosulfide and sulfur is first order with k = 2.80 ✕ ✕ 10−7 sec-1 at 1
madam [21]

Answer: a) 1.97 grams of carbon disulfide will remain after 37.0 days.

              b) 2.85 grams of carbon monosulfide will be formed after 37.0 days.

Explanation: The decomposition of carbon disulfide is given as:

                          CS_2(g)\rightarrow CS(g)+S(g)

at t=0                    4.83g             0          0

at t=37 days        4.83 - x            x           x

here,

x = amount of CS_2 utilised in the reaction

This reaction follows first order kinetics so the rate law equation is:

k=\frac{2.303}{t}log\frac{A_o}{A}

where, k = rate constant

t = time

A_o = Initial mass of reactant

A = Final mass of reactant

a) For this, the value of

k=2.80\times10^{-7}sec^{-1}

t = 370 days = 3196800 sec

A_o = 4.83

A = 4.83-x

Putting values in the above equation, we get

2.8\times 10^{-7}sec^{-1}=\frac{2.303}{3196800sec}log\left(\frac{4.83}{4.83-x}\right)

x = 2.85g

Amount of CS_2 remained after 37 days = 4.83 - x

                                                                     = 1.97g

b) Amount of carbon monosulfide formed will be equal to "x" only which we have calculated in the previous part.

Amount of carbon monosulfide formed = 2.85g

6 0
3 years ago
Select the correct answer. In which state of matter do molecules have the lowest kinetic energy? A. liquid state B. plasma state
Svetllana [295]

Answer:

D. Solid state

Explanation:

In solid state the particles of  a substance, vibrate in fixed positions though they posses less kinetic energy. This is the reason as to why the particles do not move far apart from each other making the solids remain in fixed shapes unless an extra energy is added.

8 0
3 years ago
Read 2 more answers
By which mechanism would a steroid molecule diffuse into the cell?
LUCKY_DIMON [66]

Answer: Directly through the phospholipid membrane

Explanation:

The cell membrane consist of a phospholipid bilayers structure. In the interior of the membrane, the phospholipid tail are hydrophobic, which makes the cell membrane to be selectively permeable, it is permeable to non polar molecules and impermeable to polar molecules.

Because Steroids are fat soluble, non polar compounds, they can diffuse directly through the hydrophobic, non polar core of the phospholipid bilayer without the use of carrier proteins.

5 0
3 years ago
A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh
Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C =  -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C =  4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0
3 years ago
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