All categories

2 years ago

The percent composition of nh42c2o4

Chemistry

1 answer:

zzz [600]2 years ago

5 0

Answer: Percent composition by element

Element Symbol Mass Percent

Hydrogen H 6.498%

Carbon C 19.357%

Nitrogen N 22.574%

Oxygen O 51.571%

HOPE THIS HELPS

You might be interested in

A scientist has 0.12 miles of a gas at a pressure of 4.06 kPa and a volume of 14 liters in an enclosed container. What is the te

natka813 [3]

Answer:

56972.17K

Explanation:

P = 4.06kPa = 4.06×10³Pa

V = 14L

n = 0.12 moles

R = 8.314J/Mol.K

T = ?

We need ideal gas equation to solve this question

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

T = PV / nR

T = (4.06×10³ × 14) / (0.12 × 8.314)

T = 56840 / 0.99768

T = 56972.17K

Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa

7 0

3 years ago

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83

guapka [62]

Answer:

a. 478.69 K

b. 939.43 $cm^{3}$

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

$V_{o}$ = 785 $cm^{3}$ = 0.000785 $m^{3}$

$T_{o}$ = 400K

$P_{o}$ = 125 Kpa = 125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 $m^{3}$ ⋅Pa⋅ $K^{-1}$ ⋅ $mol^{-1}$

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5 $T_{1}$ - (34980 + 35.5 $T_{o}$ )]

83.8 = (0.03 × 35.5) ( $T_{1}$ - 400K)

83.8 = 1.065 ( $T_{1}$ - 400)

78.69 = ( $T_{1}$ - 400)

$T_{1}$ = 400 + 78.69

$T_{1}$ = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

$V_{1}$ / $T_{1}$ = $V_{o}$ / $T_{o}$

$V_{1}$ = $V_{o}$ $T_{1}$ / $T_{o}$

$V_{1}$ = (785 × 478.69)/400

$V_{1}$ = 939.43 $cm^{3}$

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P( $V_{1}$ - $V_{o}$ ) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0

2 years ago

By which process is sediment laid down? its a test it's due at 10!

Alexus [3.1K]

Answer: Deposition

Explanation:

7 0

2 years ago

Read 2 more answers

Consider the following mechanism.

solong [7]

Answer :

(a) The overall equation is:

$ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)$

(b) The intermediates are :

$OH^-(aq),HClO(aq)\text{ and }HIO(aq)$

Explanation :

Part (a) :

(1) $ClO^-(aq)+H_2O(l)\rightarrow HClO(aq)+OH^-(aq)$ (fast)

(2) $I^-(aq)+HClO(aq)\rightarrow HIO(aq)+Cl^-(aq)$ (slow)

(3) $OH^-(aq)+HIO(aq)\rightarrow H_2O(l)+IO^-(aq)$ (fast)

By adding the three equations and cancelling the common terms on both side, we will get the overall equation.

$ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)$

Part (b) :

Intermediates are generated and consumed in the mechanism and do not include in the overall equation.

Since, intermediates will not include in the overall mechanism.

The intermediates are :

$OH^-(aq),HClO(aq)\text{ and }HIO(aq)$

7 0

3 years ago

By sprouting in the rainforest canopy, the strangler fig seedling is exposed to more

stiks02 [169]

Answer:

Waxy leaves protect the strangler fig from drying winds and sunlight that it is exposed to high in the canopy. Perhaps the most amazing part of this extraordinary tree is its flower. What we think of as the fruit is really a hollow, flower-bearing structure called a cyconia.

8 0

2 years ago