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Neporo4naja [7]

2 years ago

13

A gas sample contained in cylinder equipped with a moveable piston occupied 300.0 mL at a pressure of 20.0 atm. what would be th

e final pressure if the volume were increased to 500.0 mL at constant temperature?

1 answer:

7nadin3 [17]2 years ago

6 0

P1 × V1 = P2 × V2
20 × 300 = P2 × 500
6000 = P2 × 500
12 = P2

The final pressure is 12.0 atm.

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A atom has 20 electrons, 21 neutrons, and 20 protons. What is the atomic mass of the atom?

UNO [17]

name= calcium

atomic mass= 40.078 atomic mass unit

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3 years ago

Metals react with oxygen to give oxides with the general formula MxOy. What is a balanced chemical equation for the reaction of

katovenus [111]

Answer: $4Fe+3O_2\rightarrow 2Fe_2O_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here iron is having an oxidation state of +3 called as $Fe^{3+}$ cation and oxide $O^{2-}$ is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $Fe_2O_3$ .

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$2Fe+3O_2\rightarrow 2Fe_2O_3$

8 0

3 years ago

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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

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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>

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3 years ago

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A beta emission radioactive decay can pass through the body.

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3 years ago

PLEASE HELP DUE SOON

jonny [76]

Answer:

This is all true if the atom has to be neutral.

Also what does V mean?

Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.

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Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.

Neon: two shells with 10 neutrons and 10 protons in the center, with 2 electrons in the first shell, and 8 electrons in the second shell.

Boron: two shells with 6 neutrons and 5 protons in the center, with 2 electrons in the first shell, and 3 electrons in the second shell.

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2 years ago