Answer: Temperature is an example of a quantitative variable
Explanation:
A quantitative variable is defined as :
- A variable that can assume a numerical value .
- It can be ordered with respect to either magnitude or dimensions.
- It is further classified into two types : interval scale and ratio scale.
Temperature comes under interval scale , because interval scale has no zero point.
For example : A 0° C Celsius does not interpret that there is no temperature.
Therefore , Temperature is an example of a quantitative variable.
Hence, the correct answer is "quantitative variable"
Surveys are the data collection tool and research methods. The questionnaire survey tool is used to collect the data based on questions.
<h3>What are questionnaires?</h3>
Questionnaires are the tool used for the research and the data collection for the analysis, investigation, and conclusion. It is one of the data collection tools that comprises questions and prompts.
Questionnaires are the part of surveys that gathers information on the topic of the research and is easy to analyze and cost-effective. It is a valuable tool that collects data from a large number of participants.
Therefore, the correct blank is the <u>questionnaire</u>.
Learn more about the questionnaire here:
brainly.com/question/20335282
#SPJ1
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
Na 2Co3
1*23=23. 2*12=24. 6*16=96
23+24+96=143
(23*100)/143. (24*100)/143. (96*100)/143
=16%. =16.7%. =67.1%
