The Northern Hemisphere is _ during the _.

Chemistry

2 answers:

Nana76 [90]3 years ago

8 0

The Northern Hemisphere is dark during the winter.

enyata [817]3 years ago

5 0

The Northern hemisphere is tilted toward the sun, and during that time, the Southern Hemisphere is tilted away from the sun.

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The spot on the fault where the pressure first releases is called the

anyanavicka [17]

Answer:

This movement releases energy and generates seismic waves that can be recorded by specialized instruments used by scientists. The point on a fault at which the first movement or break occurs during an earthquake is called the earthquake's hypocenter

Explanation:

6 0

3 years ago

What is the total number of atoms in this formula? NaHCO3

Pavel [41]

6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.

7 0

4 years ago

60 POINTS + BRAINLIEST !! PLEASE HELP!!

uranmaximum [27]

To get moles. divide mass by molar mass.Molar mass of
Na is 23
and for Cl is 35.5.
the total molar mass of NaCl is 23+35.5 = 58.5mol/gUse the mass and divide by this number30.22g divide by 58.5mol/g and you will get 0.5166 mole.
Since the molecule has 1 Na to 1 Cl, and that the number of moles for NaCL is 0.5166. All of them would be 0.5166molesNa = 0.5166 x 1 = 0.5166molesCl = 0.5166 x 1 = 0.5166moles
to get number of atoms. Multiply your mole by Avogadro number which is 6.022x10^23Na = 0.5166 x 6.022E23 = 3.111x10^23Cl = 0.5166 x 6.022E23 = 3.111x10^23

5 0

4 years ago

Calculate the amount of energy required to increase the temperature of 18.0 go vapor water from 275.0 degrees * C to 385 degrees

Archy [21]

Answer:

The amount of energy required to increase temperature 275°C to 385°C is ΔQ =3.95208kJ

Explanation:

Given :

The mass of the water vapour is 18 g
The initial temperature : 275°C
The final temperature : 385°C

According to the formula :

Δ $Q = m*s*$ Δ $T$

Where ,

Δ $Q$ is the heat supplied
$m$ is the mass of the substance (water vapour) = 18g
$s$ is the specific heat of the substance ( water vapour) = $1.996Jg^{-1}K^{-1}$
Δ $T$ is the change in temperature = $385-275=110K$