Coastal waves dissolving limestone sediments

CaHeK987 [17]

1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
or simpler: m(N₂) = 0,225 g + 0,5331 - 0,3114 g = 0,4467 g.

2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
n - amount of substance

8 0

3 years ago

1. Define action energy.

WARRIOR [948]

Answer:

In physics, action is an attribute of the dynamics of a physical system from which the equations of motion of the system can be derived through the principle of stationary action. ... Action has dimensions of energy⋅time or momentum⋅length, and its SI unit is joule-second.

Explanation:

hope it helps

6 0

3 years ago

Read 2 more answers

A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250

Crank

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

$n=0.00250 M\times 0.002 L=0.000005 mol$

B. 5.00 mL of 0.00250 M $KSCN$

Moles of KSCN = n'

Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

$n'=0.00250 M\times 0.005 L=0.0000125 mol$

C. 3.00 mL of 0.050 M $HNO_3$

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

Concentration of nitric acid :

$[HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M$

Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

$Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}$

given concentration of $Fe(NCS)^{2+}$ at equilbrium = $3.6\times 10^{-5} M = 0.000036 M$

initially :

0.0005 M 0.00125 M 0

At equilibrium

(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M

0.000464 M 0.001214 M 0.000036 M

The expression of an equilibrium constant will be given as;

$K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}$

$=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91$

The value for the equilibrium constant is 63.91.

6 0

3 years ago

47. A substance has a half life of 3min. After 5 min, the

oee [108]

Answer:

1000

Explanation:

PLS GIVE BRAINLIEST

3 0

4 years ago

A protostaar forms once the nebular cloud condenses and the core begins

irga5000 [103]

The protostar forms once the nebular cloud condenses and the core begins to hear.

3 0

2 years ago