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2 years ago

NEED HELP PLEASE HURRY

Chemistry

1 answer:

baherus [9]2 years ago

7 0

Answer:

I believe it would be the last 2

Explanation:

the 4th one is magnesium and something else but that's with decomposition and so is the last one. the first three has water and oxygen and stuff that doesn't rly do decomposition

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If I touch a hot object, my hand gets hot through which type of heat transfer?

mamaluj [8]

Answer:

Thermal Transfer.

Explanation:

4 0

3 years ago

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A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is sub

stepladder [879]

<u>Answer:</u> The new water level of the cylinder is 24.16 mL

<u>Explanation:</u>

To calculate the volume of water displaced by silver, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

$10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL$

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

5 0

3 years ago

Help me please thanks

Evgesh-ka [11]

Then answer would be D. Answer D is correct because you would need to use a better solvent to see the ink separate on the chromatography paper. Hope that helps. :)

6 0

3 years ago

First, I would balance the chemical equation.

andrezito [222]

......................ok

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3 years ago

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Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3

jeyben [28]

Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa

8 0

3 years ago