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3 years ago

Please somebody help I have no idea what the correct answer is! (iii) and (iv)

Chemistry

1 answer:

brilliants [131]3 years ago

3 0

Since phosphoric acid is H3PO4, which is known from PO4, with a charge of 3- so 3 hydrogen would balance it out, and sodium hydroxide is NaOH, it can be assumed that it results in H3(OH)3 + Na3PO4.

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A device that sends electrical impulses to the heart to correct an irregular

IgorC [24]

A. A pacemaker

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2 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago

A third baseman throws a ball from third to first. The ball travels a 20 m/s of 2 seconds. What is the distance between third ba

umka21 [38]

Answer:

the answer would be D

Explanation:

5 0

2 years ago

Is sulfur a metal, nonmetal, metalloid, or noble gas?

Blababa [14]

Sulfer is a nonmetal

6 0

1 year ago

Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch

geniusboy [140]

F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2

3 0

2 years ago