Process in which water vapor turns to liquid

1 answer:

Rudiy274 years ago

6 0

In the amtmosphere this water vapor cools and turns back to liquid for forming clouds
this process is called condensation. When it cant hold on much longer it will turn into rain or snow.

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What is the total number of protons in an atom with the electron configuration 2-8-18-32-18-1.

svp [43]

The total number of protons in this atom is 79 because an atomic number of an element is equal to the number of protons. Here, the atomic number is 79 after adding the given electronic configuration.

What are Protons?

Every atom has a proton, a subatomic particle, in its nucleus. The particle possesses an electrical charge that is positive and opposite to the electrons.

What is Atom?

A nucleus plus one or more electrons bound to the nucleus make up an atom. The quantity of protons or electrons in an element's atom determines how different it is from other similar elements. The atomic number of an element, which serves as its primary identification, is the sum of its protons or electrons.

What is Electronic configuration?

The arrangement of electrons in atomic or molecular orbitals within an atom, molecule, or other physical structure is known as the electron configuration.

Hence, the total number of protons in this atom is 79, after adding 2 + 8 + 18 + 32 + 18 + 1.

To know more about Atom, check out:

brainly.com/question/6258301

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5 0

1 year ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>