Why is it important to read over the SDS sheet before you do a lab?

Chemistry

1 answer:

11Alexandr11 [23.1K]4 years ago

6 0

Answer:

so you know what to do and you know the rules

Explanation:

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Chemical formula for Potassium Hydrogen Nitrate

My name is Ann [436]

Answer:

KNO3

Explanation:

8 0

3 years ago

A metal object has a mass of 8.37g. When it was in a graduated cylinder containing 20.0 ml of water, the water level rose to 23.

Yuri [45]

Mass of metal is 8.37 g, volume of water in graduated cylinder is 20 mL. After adding the metal, water level increases to 23.1 mL. The volume of metal is equal to the increase in water level.

The increase in water level can be calculated by taking difference of final and initial level of water.

Thus,

$V_{metal}=V_{f}-V_{i}=(23.1-20)mL=3.1 mL$

Density of metal is defined as mass per unit volume of metal. It can be calculated as follows:

$d=\frac{m}{V}=\frac{8.37 g}{3.1 g/mL}=2.7 g/mL$

This value corresponds to the density of aluminum.

Therefore, density is 2.7 g/mL and metal is aluminum.

3 0

4 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$