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Marrrta [24]
3 years ago
13

Whta gasses are in Neptune?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
6 0
Neptune's atmosphere is composed up of many gases. These gases include Hydrogen (80%), Helium (19%), and Methane<span> (1.5%). Some of the minor gases include trace amounts of </span>Hydrogen Deuteride<span> and </span>Ethane<span>.</span>
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Match each of the following terms with the correct definition. Question 6 options: the variable that is measured as data in an e
Aleksandr [31]

The variable that is measured as data in an experiment is 1. the dependent variable.

The variables that are held constant in an experiment are 2. the controlled variables.

The variable that is changed by the experimenter is 5. the independent variable.

A count or measurement recorded during an experiment is 4. quantitative data.

Descriptions or observations during an experiment are 3. qualitative data.

5 0
3 years ago
Read 2 more answers
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0
3 years ago
The pen on seismograph swings freely.
Sever21 [200]
This statment is false, the pen on a seismograph does not swing freely.
7 0
3 years ago
This part insulates the reaction chamber from the transfer of heat to or from the surrounding environment. this part contains a
Masja [62]

The amount of heat transferred in and out of the system is measured by calorimetry. The thermometer in the calorimeter is used to measure the temperature.

<h3>What are the parts of the calorimetry device?</h3>

The thermometer (A) is a device used to measure the final and the initial temperature of the water or any other liquid in a system. A metal vessel is a place where the reaction mixture is present.

In-vessel (B), water, and metal are placed before the beginning of the experiment. The styrofoam cup or the outer metal vessel (C) insulates the instrument, from regulating the heat transformation.

Therefore, part A measures the temperature of the reaction mixture.

Learn more about insulated containers here:

brainly.com/question/866735

3 0
2 years ago
What is the volume of 300. g of mercury vapor at 822K and 0.900 atm?
Dominik [7]

Answer:

112.2L

Explanation:

Volume (V) = 300g

Temperature (T) = 822K

Pressure (P) = 0.9atm

using the ideal gas equation;

PV = nRT\\\\ V = \frac{nRT}{P}

Molar gas constant (R) = 0.0821L.atm/mol.K

Mole (n) = \frac{Mass (m)}{Molar mass (M)}                Molar mass of Mercury  = 200.59g/mol

n = \frac{300g}{200.59 g/mol} \\

   = 1.496mol

Now, the volume can be calculated;

V = \frac{1.496mol* 0.0821L.atm/mol.K*822K}{0.9atm}

∴Volume of mercury = 112.2L

8 0
2 years ago
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