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2 years ago

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Jshsbdbfbbbfffbbdvdvsgssggdhdh

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11Alexandr11 [23.1K]2 years ago

8 0

Very nice english.......

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18.2L of gas at 95°C and 760 torr is placed in a 15L container at 80 degrees * C ; what is the new pressure ?

romanna [79]

Answer:

884.56 torr

Explanation:

Formula: $\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}$

P = Pressure

V = Volume

T = Temperature in kelvin (Celsius + 273.15)

$\frac{(760)(18.2) }{368.15} = \frac{P(15) }{353.15}$

$P = \frac{(760)(18.2)(353.15) }{(368.15)(15)}$

P = 884.56169

4 0

2 years ago

Which of the following is the strongest base? ta) CH,ONa (b) NaNH, (c) CH-CH,Li (0) NaOH (6) CHÚCONa

Mashutka [201]

Hey there:

Correct answer is :

(b) NaNH₂

Sodium azanide NaNH₂ is the conjugate base of ammonia NH₃

Correct answer is :

(b) NaNH₂

I hope this will help !

7 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

evablogger [386]

Answer:

!atoms in the nitrogen family.. have 5 valence electrons. They tend to share electrons when they bond. Other elements in this family are phosphorus, arsenic, antimony, and bismuth.

5 0

2 years ago

What is the mole ratio of PCl3 to PCl5?

Ray Of Light [21]

It 1:1 I had an online review and thats the answer it told me it was.
-Hope this helped you

8 0

2 years ago

Read 2 more answers