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3 years ago

What is oxydation number of s for H2S ? ( hydrogen sulphide )

Chemistry

1 answer:

siniylev [52]3 years ago

4 0

Answer:

the oxidation number is zero because H=+1. S=-2

Explanation:

so two hydrogen and one sulphur so total no is zero

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Property of water?? I have no idea

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4 years ago

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If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t

Thepotemich [5.8K]

Answer: The final temperature of the mixture is 49°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g$

For hot water:

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g$

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 120 g

$m_2$ = mass of cold water = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of hot water = 95°C

$T_2$ = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=49^oC$

Hence, the final temperature of the mixture is 49°C

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3 years ago

the specific heat of iron is 0.4494 j/g°C. how much heat energy (kj) is transferred when a 24.7 kg iron ingot is cooled from 880

guapka [62]

Answer is: 9623.85 kJ of heat is transferred from iron ingot.
m(Fe) = 24.7 kg · 1000 g/kg = 24700 g; mass of iron ingot.
C = 0.4494 J/g°C; the specific heat of iron
ΔT = 880°C - 13°C; temperature difference.
ΔT = 867°C.
Q = m·C·ΔT.
Q = 24700 g · 0.4494 J/g°C ·867°C.
Q = 9623856.06 J ÷ 1000J/kJ.
Q = 9623.85 kJ.

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3 years ago

What is oxydation number of s for H2S ? ( hydrogen sulphide )​

What is oxydation number of s for H2S ? ( hydrogen sulphide )