All categories

3 years ago

4. If an experimental value for a given observation was 415 nm while the theoretical value

Chemistry

1 answer:

Alex3 years ago

7 0

Answer: % error of observation is 4.77%

Explanation:

Given:

Observation value = 415nm

theoretical value= 435.8nm

Percent error of observation = theoretical value- observation value/ theoretical value x 100 %

= 435.8-415/435.8= 0.04772 x 100 = 4.77%

therefore % error of observation is 4.77%

You might be interested in

Somebody help with A B and C please will give brain thanks

Norma-Jean [14]

Answer:

we cant see a b or c

Explanation:

7 0

2 years ago

Where does most of the mass of an atom come from?

Sedaia [141]

Where does most of the mass of the universe come from? In ordinary matter, most of the mass is contained in atoms, and the majority of the mass of an atom resides in the nucleus, made of protons and neutrons. Protons and neutrons are each made of three quarks.

7 0

3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

What is another name for hard water?

astra-53 [7]

Answer:

Hyponym is the another name for hard water.

Explanation:

Please Mark me brainliest

7 0

2 years ago

The standard unit (SI) for mass is

telo118 [61]

The kilogram is the si

6 0

3 years ago