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Softa [21]
3 years ago
10

According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?

Chemistry
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

C. paramagnetic.

Explanation:

1s2 2s2 2p6 3s2 3p6 4s2 3d3 It is Vanadium.

It has 3 unpaired d- electrons, so it is paramagnetic.

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Name two uses for seismometers
NNADVOKAT [17]
Earthquakes and volcanic eruption
8 0
3 years ago
The estimated costs for remodelling the interior of an apartment are: three 1-gallon cans of paint at $15.44 each, two paint bru
Vlad1618 [11]

The total estimated cost with the appropriate significant figures is

$195,13.

Any two non-zero digits that are separated by a zero are important. For instance, the number 108.0097 has seven significant digits. Every zero that is both to the right and left of a non-zero digit and the decimal point is never meaningful. For instance, the number 0.00798 had three significant digits.

Simply multiply the required number of goods by their individual costs, then add them all up to calculate this:

3*13,69= 41,07 2*9,53=19,06 \s1*135= 135

Total cost: $41,07 plus $19,06 plus 135 equals $55,13.

Thus, the overall price will be $195,13.

Learn more about significant figures here-

brainly.com/question/14359464

#SPJ9

7 0
2 years ago
The chart below shows examples of solids, liquids, gases, and plasma.
spin [16.1K]
The word that is in an incorrect place in the chart is HAIL.
The state of matter of hail is not plasma; hail is a solid at room temperature. Plasma refers to the state of matter in which an ionized gas have approximately equal number of positively and negatively charged ions. Thus, plasma is an ionized gas. Plasma is considered to be the fourth state of matter.  <span />
4 0
3 years ago
Read 2 more answers
Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small a
GrogVix [38]

Answer:

The calorie content of  65g of candy is 326.78 cal

Explanation:

Step 1: Data given

Mass of the candy = 15.00 grams

Mass of the container = 0.325 kg

Mass of water = 1.75kg

0.624 kg at an initial temperature of 15.0°C.

The specific heat of aluminium = 0.22 Cal/kg°C

The specific heat of water = 1 cal/kg°C

Step 2: Calculate calorie content for a 15 gram sample

ΔQ = Σm*c*ΔT

 ⇒ m = mass in grams

⇒ with c= the specific heat in Cal/kg°C

⇒ with ΔT = T2 -T1 = the change in temperatures in °C

ΔQ = m(bomb) * C(aluminium) * ΔT + m(cup) * C(aluminium) * ΔT + m(H2O) * c(H20) * ΔT

ΔQ = (m(bomb) + m(cup)) * c(aluminium)  + m(H2O)*c(H20) ) * ΔT

⇒ with mass of the bomb calorimeter = 0.325 kg

⇒ with mass of the cup = 0.624 kg

⇒ with c(aluminium) = the specific heat of aluminium = 0.22 Cal/kg°C

⇒ with mass of water = 1.75 kg

⇒ with c(water) = the heat capacity of water = 1 Cal/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 53.5 - 15.0 = 38.5 °C

ΔQ = 0.325*0.22*38.5 + 0.624*0.22*38.5 + 1.75*1*38.5

ΔQ = ((0.325 + 0.624)*0.22 + 1.75*1)*38.5

ΔQ = 75.41 cal

Step 3: Calculate the calorie content for a 65 gram sample

For a 65g sample the calorie content will be more or less 4x higher than a 15 gram sample:

ΔQ = 75.41  * (65/15) = 326.78 cal

8 0
3 years ago
If you were about 10 kilometers above earth, what would the pressure approximately be?
Sliva [168]

Answer:

About 25 kPa  

Explanation:

Pressure decreases with height above sea level.

The calculation is rather complicated, so I will refer to the figure below.  

It shows that the pressure at 10 km is about 25 kPa.

7 0
3 years ago
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