In what type of solids are the particles not arranged in a regular pattern

GarryVolchara [31]

1) (C2H5)2CBrCH2CH3 is the answer

explaiation:-

so when HBr is added to an alkene , according to the Markonicoff's rule ...H atoms are bonded to the C containing the most amount of H and Br is added to the other C.

2) Just add alkoholic KOH∆

6 0

3 years ago

What is decarboxylation?? Gimme one reaction of it..

kotegsom [21]

Answer:

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.

Explanation:

Please give me brainlist

3 0

3 years ago

Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is th

Pachacha [2.7K]

Answer:

5.004kg

Explanation:

Combustion of carbon

C+O2=CO2

from the relationship of molar ratio

mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)

mass of carbon =1000kg

atomic mass of carbon =12

volume of CO2 produced=1000×22.4/12

volume of CO2 produced =1866.6dm3

from the combustion reaction equation provided

CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

applying the same relationship of molar ratio

no of mole of CO2=no of mole of urea

therefore

vol of CO2\22.4=mass of urea/molar mass of urea

molar mass of urea=60.06g/mol

from the first calculation

vol of CO2=1866.6dm3

mass of urea=1866.6×60.06/22.4

mass of urea=5004.82kg

7 0

3 years ago

Read 2 more answers

What is the volume of 2 mol of chlorine gas at STP? 2.0 L 11.2 L 22.4 L 44.8 L

nirvana33 [79]

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

7 0

2 years ago

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago