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Zepler [3.9K]
3 years ago
10

QUICK CHECK

Chemistry
2 answers:
allochka39001 [22]3 years ago
6 0

Answer:

Tetraedral: CH4

Bent: H2O

Trigonal planar: BF3

Trigonal pyramidal: NH3

Linear: CO2

Explanation:

edge

Free_Kalibri [48]3 years ago
5 0

Answer:

Tetraedral: CH4

Bent: H2O

Trigonal planar: BF3

Trigonal pyramidal: NH3

Linear: CO2

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You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the
lara31 [8.8K]

Answer:

m_{ice} = 65.336\,g

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

Q_{water} = Q_{ice}

(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)

The amount of ice that is melt is:

m_{ice} = 65.336\,g

5 0
4 years ago
Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h
balu736 [363]

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

                             .430mol  x 12  =  5.16 g C

                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0
4 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
Looking at the Bohr models of atoms offer an explanation as to WHY the atomic size changes as it does for both periods and group
andriy [413]
Element 87 francium have the highest atomic size, so in short the closest element to francium have the high atomic size.

5 0
3 years ago
At equal concentrations would a nonelectrolyte (e.g. glucose) or electrolyte (e.g. nacl) containing solution have a lower freezi
Anestetic [448]
Answer is: solution of electrolyte will have lower freezing point than solution of nonelectrolyte.
This is because salt solution has more particles in of sodium chloride (sodium and chlorine ions) than in same concentration of glucose. Electrolytes better separates into particles in water because of their ionic bond.<span>
</span>
6 0
3 years ago
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