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Marysya12 [62]
3 years ago
9

Clinical thermomerter​

Chemistry
2 answers:
Yuki888 [10]3 years ago
6 0
A small medical thermometer with a short but finely calibrated range, for taking a person's temperature.
goldenfox [79]3 years ago
3 0

Answer:

Here is the definition: A medical thermometer (also called clinical thermometer) is used for measuring human or animal body temperature.

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How do you determine the correct subscripts in a chemical formula
VMariaS [17]

BY CHECKING THE REACTIVITY OF AN ELEMENT WHICH IS MOST REACTIVE OR NOT AND YOU STUDY TYPE OF CHEMICAL REACTION IN 1 CH AND YOU CHECK THE REACTIVITY OF ELEMENTS IN 3 CH METALS AND NON METALS PAGE NO 45 IN NCERT BOOK

4 0
3 years ago
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
The element with the most stable nucleus and smallest mass per particle is
Margaret [11]
"18) the element with the most stable nucleus and smallest mass per particle is:A) uranium.B)argon C) helium."
4 0
3 years ago
What is the ratio of hydrogen atoms to sulfur atoms in sulfuric acid, h2so4?
Vsevolod [243]
I think it's 2:1 or 2:1:4. I mostly think it's 2:1 though. (:
4 0
3 years ago
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
seraphim [82]

Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
3 years ago
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