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3 years ago

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Chemistry

1 answer:

liberstina [14]3 years ago

3 0

Answer: 1-Exhale 2-inhale 3-contracted diaphragm 4-relaxed diaphragm.
Step by step explanation: hope this helped! Have a great day.

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2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme

mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

$2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O$

Next, we assign the oxidation numbers as follows:

$2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}$

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0

3 years ago

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dedylja [7]

Um all? i’m confused

5 0

3 years ago

How many inches are in 25.8 cm

NeTakaya

Answer:

10.1574803 inches

Explanation:

Here you go

6 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Trava [24]

Answer:

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4 0

3 years ago

Read 2 more answers

Anybody good at chemistry?

Luba_88 [7]

Answer:

Explanation:

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C - 40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C × 55°C

Q = 42.35 j

Given data:

Mass = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C - 20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C × 80°C

Q = 95.48 j

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C - 60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C × -30°C

120 j = m × -26.91 j/g

m = 120 j / -26.91 j/g

m = 4.46 g

negative sign show heat is released.

Given data:

Mass of ice = 1.5 g

Change in temperature = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C × ΔT

30.0 j = 0.753 j/°C × ΔT

30.0 j /0.753 j/°C = ΔT

39.84 °C = ΔT

3 0

3 years ago