Answer:
9 (1-2x²)
Explanation:
The given expression is:
30 - 9x²*2 - 21 - 4 + 4
The first step is to compute the multiplication. This will give:
30 - 18x² - 21 - 4 + 4
Then, we will add like terms as follows:
(30-21-4+4) - 18x²
= 9 - 18x²
Finally, we can take the 9 as a common factor from both terms, this will give:
9 (1-2x²)
Hope this helps :)
This statement is True! Lets think about it... When water boils, the water doesnt evaporate from the bottom of the bowl it evaporates from the top!
=)
The one with higher temperature is the one with NaOH as heatis given off during the neutralization reaction that occurs.
<h3>
What is volume?</h3>
Volume can be defined as the amount of space a substance or an objects occupies usually in a closed container.
Volume is measures in litres.
When water is added to dilute acid like HCl, they become more dilute.
When NaOH is added to HCl, a neutralization reaction occurs.
The student will determine the contents of the flasks by adding 10 ml of hcl to each flask. If the NaOH reacts with the Hcl, there will be an increase in temperature.
The increase in temperature is due to the heat of neutralization of the reaction between NaOH and HCl.
Learn more about volume at: brainly.com/question/1972490
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<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Answer:
Towards this goal, this project aims to develop a statistical measure of the uncertainty of the decisions made on the friction ridge evidence (i.e., evidential value of fingerprint comparison), which ultimately can be referred to as a scientific basis of the identification decisions made in friction ridge analysis.
Explanation:
