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Sauron [17]
3 years ago
5

the elements carbon hydrogen and oxygen are all part of the same blank on the periodic table........ i i think GROUP diangol row

period
Chemistry
1 answer:
Mrrafil [7]3 years ago
6 0
The elements Carbon, Hydrogen, and Oxygen are all part of Non-Metals.

Hope This Helps! :)
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Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas:NaH (s) H2O (l) → NaOH (aq) H2 (g)A
Anuta_ua [19.1K]

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

You want to calculate the mass of NaH, I assume.  Otherwise, the question isn't clear.  It simply says calculate the mass(??)

 

So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr.  But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas.  So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.

PV = nRT

n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)

n = 0.0385 moles H2

 

moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required

mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH

Brainliest Please :)

7 0
3 years ago
Read 2 more answers
What is the molarity of a solution that contains 14 moles of solute and 2 liters of solution?
Nata [24]

Answer:

  7 M

Explanation:

The molarity is defined as "moles per liter." Here, it is ...

  (14 moles)/(2 liters) = 7 moles/liter = 7 M

6 0
3 years ago
How many moles of mercury are equivalent to 3.46 x 1023 atoms?
olga55 [171]

Answer:

3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

4 0
3 years ago
Pls help
Andrej [43]

Answer:

• None of these

Explanation:

your hand looks too old.

#Rastaman

5 0
2 years ago
How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

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

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0
3 years ago
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