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TiliK225 [7]
3 years ago
5

What's a beaker with a hole in it where a tube can be inserted to be connected to another beaker called?

Chemistry
1 answer:
Darya [45]3 years ago
8 0

Answer:

its in g,oogle

Explanation:

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<img src="https://tex.z-dn.net/?f=what%20%5C%3A%20is%20%5C%3A%20acid%20%5C%3A%20%20%20%5C%3A%20%20%5C%3A%20%7B%3F%7D%20" id="Tex
tensa zangetsu [6.8K]

meaning:An acid is a chemical substance, usually a liquid, which contains hydrogen and can react with other substances to form salts. Some acids burn or dissolve other substances that they come into contact with.

5 0
3 years ago
Read 2 more answers
vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compa
Vesna [10]

Answer:

q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.

Explanation:

1. Heat absorbed

q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ

2. Change in volume

V(water) = 0.018 L

pV = nRT

1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K

V = 30.62 L

ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L

3. Work done

w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm

w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ

4. Why the difference?

Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.

The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.

7 0
3 years ago
Necesito ayuda porfavor
AlexFokin [52]

Answer:

thank for the points ❤️

Explanation:

pa breinlyest po

8 0
3 years ago
Classify CH3CH2NHCH3 as a strong base or a weak base
harina [27]

CH3CH2NHCH3 is a weak base.

5 0
3 years ago
Read 2 more answers
At a temperature of -25 °C, a sample of gas in a rigid container exerts a pressure of 55.8 kPa. At what temperature will the pre
never [62]
<h3>Given:</h3>

P_{1} = \text{55.8 kPa}

T_{1} = -25^{\circ}\text{C + 273 = 248 K}

P_{2} = \text{145 kPa}

<h3>Unknown:</h3>

T_{2}

<h3>Solution:</h3>

\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}

T_{2} = T_{1} \times \dfrac{P_{2}}{P_{1}}

T_{2} = \text{248 K} \times \dfrac{\text{145 kPa}}{\text{55.8 kPa}}

\boxed{T_{2} = \text{644 K}}

4 0
3 years ago
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