How do computers play a role in meteorology? Select three options.

The volume occupied by 0.25 mol of sulfur dioxide at rtp

Vladimir79 [104]

Answer:

V = 0.25 x 22.4 Liters

Explanation:

1 mole of any gas occupies 22.4 Liters at STP.

8 0

3 years ago

Which of the following would be expected to have the highest viscosity? CH3CH2OH, HOCH2OH, CH3CH2CH3

zaharov [31]

Answer: Out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

Explanation:

The resistance occurred in the flow of a liquid substance is called viscosity.

More stronger is the intermolecular forces present in a substance more will be its resistance in its flow. Hence, more will be its viscosity.

For example, $HOCH_{2}OH$ has strong intermolecular hydrogen bonding than the one's present in $CH_{3}CH_{2}OH$ and $CH_{3}CH_{2}CH_{3}$ . This is because two-OH groups are present over here.

Thus, we can conclude that out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

5 0

3 years ago

Write the atomic symbol for the isotope of bismuth with 125 neutrons.

vlada-n [284]

First, isotopes are the atoms of a single element whose nuclei have a different number of neutrons, and therefore, differ in mass numbers. You should know that atoms are formed by a nucleus that has a small size and is made up of protons and neutrons. The nucleus is surrounded by a cloud of electrons, which are found in a region of the atom called the cortex.

The mass number, represented as A, is the sum of the number of protons and neutrons in the nucleus. On the other hand, the atomic number (Z) is the number of protons that exist in the nucleus.

The isotopes of an element X are represented as follows,

(see first attached picture)

It should be noted that the number of neutrons of a chemical element can be calculated as the difference A-Z.

The atomic and mass numbers of bismuth with 125 neutrons are:

Z = 83

A = 83 + 125 = 208

Thus, the atomic symbol of the bismuth isotope with 125 neutrons is:

(see second attached picture)

5 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t

solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

6 0

3 years ago