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4 years ago

If beaker tongs are not available, which could you

Chemistry

2 answers:

Alenkasestr [34]4 years ago

4 0

<u>Answer</u>:

If beaker tongs are not available, we could use crucible tongs to handle the hot beaker

<u>Explanation</u>:

Crucible tongs is tool shaped like a scissor where instead of two blades, two pincers or pieces of metals that concave together is used. They help us to hold or grasp a hot beakers, flasks or even small beakers. It is difficult for the people to catch or hold the hot beakers directly. So crucible tongs plays a vital role when burning, or using hot objects in the chemical laboratory.Crucible tongs are generally made with strong metal such as nickel, stainless or oxidized steel so as to withstand very high and extreme temperatures.

choli [55]4 years ago

4 0

Answer:

heat resistance gloves

Explanation:

answer on edg

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3 years ago

The density of water at 400C is 0.992 g/mL What is the volume of 27.0 g of water at this temperature?

pantera1 [17]

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Volume of water at this temperature is 27.2 mL

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4 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 $\frac{cal}{gC}$ .

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.
C is the specific heat substance.
ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

Mass of metal = 50 g
Initial temperature of metal= 45 °C
Final temperature of metal= 11.08 ºC
Specific heat of metal= ?

For water:

Mass of water = 250 g
Initial temperature of water= 10 ºC
Final temperature of water= 11.08 ºC
Specific heat of water = 1.035 $\frac{cal}{gC}$

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 $\frac{cal}{gC}$ × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 $\frac{cal}{gC}$ × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= $\frac{279.45 cal}{1696 gC}$