All categories

3 years ago

24H2S + 16HNO3 3Sg+ 16NO + 32H2O

Chemistry

1 answer:

jeka943 years ago

3 0

Answer:

24 sultur with 48hydrogen+16 hydrogen,nitrogen and 48 oxygen+16 nitrogen,16 oxygen+62hydrogen and 32 oxygen

Explanation:

You might be interested in

Plz answer this i will mark brainly like and ratr

matrenka [14]

Answer:

answer what ?

Explanation:

4 0

2 years ago

Find the formula of the ionic compound made from aluminum (Al) and oxygen (O).

FrozenT [24]

The formula will be AI2O3

4 0

3 years ago

Calculate the average atomic mass of element X using the table below. Then use the periodic table to identify the element. Separ

Ksenya-84 [330]

Answer:

126.8, Iodine

Explanation:

mass ×abundance/100
(126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
102.1+21.7+3=126.8

IODINE has an atomic mass of 126.8 or 126.9

4 0

2 years ago

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

5 0

3 years ago

Read 2 more answers

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0

3 years ago