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3 years ago

What are prevailing winds?

Chemistry

1 answer:

Tems11 [23]3 years ago

7 0

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

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What is the answer to the net ionic equation FeO(s)+2HClO4(aq)--> Fe(ClO4)2 (aq)+ H2O(l)

yan [13]

Feo + 2H = H2O + Fe + 2 + CIO4-
this is ur answer. .
mrk me as brainlist

8 0

3 years ago

Read 2 more answers

Temperature is the measurement of how hot or cold something is. Agree Disagree

Nataliya [291]

Answer:

Agree

Explanation:

7 0

3 years ago

Read 2 more answers

Perform the following conversion: 6.1 × 103 K (the surface temperature of the Sun) to °F and °C. Pay attention to the number of

Keith_Richards [23]

Answer:

$\°C=5.8x10^3\°C$

$\°F=1.1x10^4\°F$

Explanation:

Hello,

In this case, for the calculation of the temperature in degree Celsius we subtract 273.15 to the given temperature in kelvins:

$\°C=6100-273.15\\\\\°C=5.8x10^3\°C$

Next, by applying the following equation we compute it in degree Fahrenheit:

$\°F=(5.8x10^{3}*9/5) + 32\\\\\°F=1.1x10^4\°F$

Clearly, since the initial unit has two significant figures the computed units also show two significant figures.

Regards.

5 0

3 years ago

A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water level rose to the 37.4 mL mark

musickatia [10]

Volume of the metal = change in volume reading
Volume = 37.4 - 33
Volume = 4.4 ml

Density = mass / volume
Density = 7.101 / 4.4
Density = 1.61 g/ml

7 0

3 years ago

Rank these transition metal ions in order of decreasing number of unpaired electrons.

lesya692 [45]

Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

Atomic number of Fe is 26.

$Fe^{3+} - [Ar] 3d^{5}$

So, there is only 1 unpaired electron present in $Fe^{3+}$ .

Atomic number of Mn is 25.

$Mn^{4+} - [Ar]3d^{3}$

So, there are only 3 unpaired electrons present in $Mn^{4+}$ .

Atomic number of V is 23.

$V^{3+} - [Ar] 3d^{2}$

So, there are only 2 unpaired electrons present in $V^{3+}$ .

Atomic number of Ni is 28.

$Ni^{2+} - [Ar] 3d^{8}$

So, there will be 2 unpaired electrons present in $Ni^{2+}$ .

Atomic number of Cu is 29.

$Cu^{+} - [Ar] 3d^{10}$

So, there is no unpaired electron present in $Cu^{+}$ .

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}$

7 0

3 years ago