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alexandr402 [8]

4 years ago

6

Part b select the ester that is formed when propanoic acid reacts with isopropyl alcohol in the presence of heat and an acid cat

alyst.

1 answer:

Otrada [13]4 years ago

7 0

<span>The ester that is formed by combining propanioc acid with isopropyl alcohol, using heat and an acid catalyst is isopropyl propanoate.</span>

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During World War II, a portable source of hydrogen gas was needed for weather balloons, and solid metal hydrides were the most c

Zolol [24]

Answer:

2758 L of LiH and 849 L of MgH₂

Explanation:

To know the volume of the gas, we can use the ideal gas law:

PV =nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (62.36 torr*L/mol*K), and T is the temperature.

The mass is 1.95 lb = 884.52 g (1lb = 453.6 g).

T = 27ºC + 273 = 300 K

<em>For lithium hydride (LiH)</em>

The molar mass is: 7 g/mol of Li + 1 g/mol of H = 8g/mol

n = 884.52/8 = 110.565 mol

750*V = 110.565*62.36 *300

V = 2758 L of LiH

<em>From magnesium hydride (MgH₂)</em>

The molar mass is: 24 g/mol of Mg + 2* 1 g/mol of H = 26 g/mol

n = 884.52/26 = 34.02 mol

750*V = 34.02*62.36 *300

V = 849 L of MgH₂

6 0

3 years ago

Which choice is an element? Air, oxygen, water, or sodium cholride

Ksju [112]

Answer:

Oxygen is an element.

Explanation:

Out of the choices listed, oxygen is the only element. The other choices are incorrect because air is a mixture of nitrogen, oxygen, carbon dioxide, water vapors, and some other gases; water is a compound of hydrogen and oxygen; and sodium chloride is a compound of sodium and chloride ions; meanwhile, oxygen is a pure element.

7 0

3 years ago

The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0

vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of $NH_3$ = 17 g/mole

Given,: 28% w/w of $NH_3$ solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml$

Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute $NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles$

$V_s$ = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.65moles}{0.11L}=15mole/L$

Using molarity equation:

$M_1V_1=M_2V_2$

$15\times V_1=0.036\times 1.0\times 10^{3}$

$V_1=2.4ml$

6 0

3 years ago

77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a

BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

Consider the sugar below. Which term best describes this sugar? disaccharide monosaccharide polysaccharide saccharide

nekit [7.7K]

oh wait i wrote whole paragraphs let me fix this

okay so you mean this ?

a term for disaccharides would be double sugar or boise!

monosaccharides are simple sugars soluble in water. any class of sugars like glucose.

polysaccharide is starch, cellulose, or glycogen. with molecules together. a carbohydrate.

saccharide knows as to the term structure of carbohydrates. carbohydrates are plain simple? organic compounds that are aldehyde or ketone functional group. with the general chemical formula of carbohydrates is c (H O)

n. 2. n

7 0

3 years ago