All categories

2 years ago

How should you remove a beaker from a hot plate after heating it?.

Chemistry

1 answer:

ohaa [14]2 years ago

3 0

Answer:

One of the students should place a wire gauze on the lab station to set the beaker on. Using the beaker tongs, one of the students should gently remove the beaker from the hot plate and set it down on to the wire gauze.

Explanation:

I hope it will help you...

You might be interested in

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

Taken at face value, does this DNA evidence incriminate Bill, Lucy, both, or neither? Explain your answer. (two to three sentenc

inysia [295]

Answer: Yes, it does incriminate both Bill and Lucy. Both had DNA found at the scene which makes them both suspects until further investigations are conducted.

Explanation: The term "at face value" means without further investigation so it is just saying looking at what has been seen are Bill and Lucy suspects. This shows that they are because both have DNA found at the scene even though they may not have been the murderer they are being investigated.

6 0

4 years ago

Select the two variables that are held constant when testing Boyle's law in a manometer.

MrRissso [65]

Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select all that apply. V2 = k/P2 V2 = P1V1/P2 ... What two variables are held constant when testing Boyle's Law in a manometer? Temperature hope this helps

4 0

3 years ago

Read 2 more answers

PLZ HELP ME WITH MY WORK

Delvig [45]

Answer:

neutrons should be added with the protons

4 0

3 years ago

1.5g of magnesium ribbon is burnt in oxygen to produce 2.5g of magnesium oxide. How much oxygen is required in grams?

Furkat [3]

Answer:

2.5 grams

Explanation:

law of conservation of mass applies. the magnesium completely burns

4 0

3 years ago