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3 years ago

Is water vapor water in the dissolved air state

Chemistry

2 answers:

cupoosta [38]3 years ago

5 0

Yes vapor is water once it gathers together it's water once more

Bess [88]3 years ago

3 0

Yes because water turns into vapor

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An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its

Aleks04 [339]

Answer:

$T_1=-91.18\°C$

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

$\frac{T_1}{V_1}= \frac{T_2}{V_2}$

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

$T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C$

Best regards!

6 0

3 years ago

Write and balance the equation for the combination of ammonia gas with solid copper (II) oxide to produce copper metal, nitrogen

Katena32 [7]

Explanation :

The balanced chemical reaction will be,

$3CuO(s)+2NH_3(g)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)$

By the stiochiometry, 3 moles of solid copper(II)oxide react with 2 moles of ammonia gas to give 3 moles of copper metal, 1 mole of nitrogen gas and 3 moles of liquid water.

The states of matter of each elements and compound is,

Copper(II)oxide is in solid state

Ammonia is in gaseous state

Copper metal is in solid state

Nitrogen is in gaseous state

Water is in liquid state

7 0

3 years ago

Read 2 more answers

What is a watt, which is a unit of power,equal to?

Bas_tet [7]

A joule times a second :)

7 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$