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WINSTONCH [101]
3 years ago
7

IUPAC name for [Fe(NH3)4Cl2]NO3

Chemistry
1 answer:
Ksju [112]3 years ago
8 0

Tetraamminedichloridoiron(3) nitrate

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A gas occupies 56 L at 73°C. What volume will the gas occupy if the temp. cools to 0°C?
vova2212 [387]

Answer:

44.2 L

Explanation:

Use Charles Law:

\frac{V1}{T1} =\frac{V2}{T2}

We have all the values except for V₂; this is what we're solving for. Input the values:

\frac{56 L}{346K} =\frac{V2}{273K}   -  make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

\frac{56*273}{346} = V2

Therefore, V₂ = 44.2 L

It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

7 0
3 years ago
P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:
mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0
3 years ago
Do bases react with metals the same way that acids do?.
mina [271]

Answer:

Bases do not react with metals in the way that acids do

Explanation:

hope this helps

pls mark brainliest

6 0
2 years ago
Identify the terms described by selecting the correct word from the drop-down menu.
Ksju [112]

Answer: Matter

Biosphere

Explanation:

7 0
3 years ago
An iron bar of mass 841 g cools from 84°C to 7°C. Calculate the heat released (in kilojoules) by the metal.
Oduvanchick [21]

Answer:

28.75211 kj

Explanation:

Given data:

Mass of iron bar = 841 g

Initial temperature = 84°C

Final temperature = 7°C

Heat released = ?

Solution:

Specific heat capacity:

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

specific heat capacity of iron is 0.444 j/g.°C

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 = 7°C - 84°C

ΔT = -77°C

By putting values,

Q = 841 g × 0.444 j/g.°C × -77°C

Q = 28752.11 j

In Kj:

28752.11 j × 1 kJ / 1000 J

28.75211 kj

8 0
3 years ago
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