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Minchanka [31]
3 years ago
5

PLEASE HELP!! ASAP!! WILL MARK BRAINLIEST!!!

Chemistry
1 answer:
sdas [7]3 years ago
7 0

Answer:

Alkali metals are soft and have low melting points.

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Two descriptions about physical quantities are given below:
gladu [14]

Answer:

Quantity A is weight and Quantity B is mass

Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object

8 0
3 years ago
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To what family of the periodic table does this new element probably belong
julsineya [31]
<span>Ionization energy (IE) is the amount of energy required to remove an electron. If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).</span>
6 0
3 years ago
A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

6 0
4 years ago
Read 2 more answers
Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?
AleksandrR [38]

This reaction would give rise to two products.

  • 2-bromo-3-methylhexane, and
  • 3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton \text{H}^{+} and a bromide ion \text{Br}^{-}.

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

  • \phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3
  • \phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon.   The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.

6 0
3 years ago
The half-life of carbon-14 is about 6000 years. assume that a sample of charcoal is formed by burning of living wood 15,000 year
zimovet [89]
The daughter isotope has an atomic number one less than the parent and a mass number two less. : A. : 1 20) The half-life<span> of </span>carbon-14<span> is about </span>6000 years<span>. </span>Assume<span> that a </span>sample<span> of </span>charcoal formed<span> by</span>burning<span> of </span>living wood 15,000 years ago<span>. How </span>much<span> of the </span>original carbon-14 would remain today? A) more than one-<span>half</span>
4 0
3 years ago
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