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2 years ago

Which in bigger 43 mg or 5 g

Chemistry

2 answers:

AleksandrR [38]2 years ago

6 0

5 g is bigger than 43 mg.

I hope this is right, I apologize if I'm wrong.

strojnjashka [21]2 years ago

5 0

5g is bigger. 43 mg is only 0.043g.

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What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3

snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

$\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol$

8 0

2 years ago

For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.

miss Akunina [59]

When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° = - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc

and Kc = 9

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

= - 4536

4 0

2 years ago

A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?

Katen [24]

V ( HCl ) = 16.4 mL / 1000 => 0.0164 L

M( HCl) = ?

V( KOH) = 12.7 mL / 1000 => 0.0127 L

M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

HCl + KOH = H2O + KCl

1 mole HCl ------ 1 mole KOH
? mole HCl--------0.007874 moles KOH

moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / 0.0164

= 0.480 M

Answer (2)

hope this helps!

4 0

2 years ago

Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3

Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

Nitrogen has 14.01 gram/mol

And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

1.01 by 3 i.e., 3.03

Now, add

14.01

+ 3.03

17.04

So, The molar mass of ammonia, NH₃ is

17.04 g/mol

-TheUnknownScientist

5 0

2 years ago

The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction

shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

$NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)$

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of $H_{2}S$ will increase.

(b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of $H_{2}S$ will decrease with decrease in volume.

When we increase the mount of $NH_{4}HS$ then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of $H_{2}S$ will increase then.

3 0

2 years ago