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3 years ago

1). How can the position of a particle be determined?

Chemistry

1 answer:

Kamila [148]3 years ago

8 0

1).C-by analyzing its interactions with another particle

2).C-18

3).C-energy sub-level

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Calculate the number of moles in 24.18 liters of neon gas at STP.

uranmaximum [27]

The number of moles of Neon gas : 1.079

<h3>Further explanation</h3>

Given

24.18 Liters of Neon gas at STP

Required

The number of moles

Solution

There are 2 conditions that are usually used as a reference in chemical calculations (mainly for determining the volume per mole of a gas or the molar volume), STP and RTP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So mol for 24.18 L :

= 24.18 : 22.4

= 1.079 moles

8 0

3 years ago

What is the difference between glucose and fructose.

CaHeK987 [17]

Answer:

glucose is in large sugar molecules, while fructose is in simple fruits and vegetables

4 0

2 years ago

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What happens if the metal you throw in is MORE REACTIVE than the<br> metal ion in solution?

IRISSAK [1]

A displacement reaction Would occur in this situation

6 0

3 years ago

Question 14 A student dissolves 1.5g of styrene C8H8 in 225.mL of a solvent with a density of 1.02/gmL . The student notices tha

Yuki888 [10]

We must to know:

Cm = molarity = niu / Vs, when the niu = no. of moles and Vs = Volume of solution

the no. niu = mass / molecular mass of substance

molecular mass of C8H8 = 12x8+8x1 = 104 g/mol

=> niu = 1,5 / 104 = 0,0144 moles C8H8

=> Cm = 0,0144/0,225 = 0,06 mol/L

Cmm = molality = niu (C8H8) / mass of solvent (kg)

=> p = mass / V => mass (solvent) = p x V

=> 225 x 1,02 = 229,5 g solvent = 0,2295 kg solvent

=> Cmm = 0,0144 / 0,229,5 = 0,063

8 0

3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$