Answer:
A. cannot tell which change produced the results.
If more than one variable is changed in an experiment, scientist cannot attribute the changes or differences in the results to one cause. By looking at and changing one variable at a time, the results can be directly attributed to the independent variable
In a place with no gravity
Explanation:
Weight is a force exerted by a body in the presence of gravity.
Weight = mass x gravity
Without gravity, there is no force of weight. The more the gravity, the more the weight of a body.
A body will experience weightlessness in a place without no gravity.
Therefore, one will experience the least weight in a place without the pull of gravity.
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I'm going to assume you are speaking about the characteristics of weather.
Here they are:
Humidity
Air Temperature and Pressure
Wind Speed/Direction
Cloud Cover and what kind of clouds.
Also the amount of precipitation as well as what effects the different kind of weather phenomenon to change into e.g. Freezing rain, Hail, Straight Line winds, etc.
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
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Answer:
2.09 atm
Explanation:
Step 1: Given and required data
- Volume of the vessel (V): 25.0 L
We won't need the data of water and uncombusted fuel, since the partial pressures are independent of each other.
Step 2: Calculate the number of moles (n) corresponding to 60.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
60.0 g × 1 mol/44.01 g = 1.36 mol
Step 3: Calculate the partial pressure of CO₂
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/ V
P = 1.36 mol × (0.0821 atm.L/mol.K) × 468.2 K/ 25.0 L = 2.09 atm
