Answer: the reliability will be worse
Explanation:
Suppose we used 0.5 M NaOH to titrate our vinegar sample instead of 0.1 M.
Now by using 0.5M instead of 0.1M we are increasing the concentration of NaOH,
We know that the moles used = Volume x concnetration.
so for the same no of moles, if the concentration increases, the volume decreases.
Hence it will consume less NaOH.
now Since the volume decreases, the titration volume of less number will increase the % error.
Therefore the reliability will be worse.
The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C
<h3>How to calculate temperature?</h3>
The initial temperature of the copper metal can be calculated using the following formula on calorimetry:
Q = mc∆T
mc∆T (water) = - mc∆T (metal)
Where;
- m = mass
- c = specific heat capacity
- ∆T = change in temperature
According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:
400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)
30,096 = 93.6T - 2246.4
93.6T = 32342.4
T = 345.5°C
Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.
Learn more about temperature at: brainly.com/question/15267055
A colloid has the particles that have the ability to scatter light called the Tyndall effect named after the scientist named Tyndall. A suspension has large suspended particles that settle out at the bottom of the container. A solution has small particles that are evenly distributed throughout. Hence the answer is choice 2.
There are 1.92 × 10^23 atoms Mo in the cylinder.
<em>Step 1</em>. Calculate the <em>mass of the cylinder
</em>
Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g
<em>Step 2</em>. Calculate the<em> mass of Mo
</em>
Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo
<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo
</em>
Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo
<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo
</em>
Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)
= 1.92 × 10^23 atoms Mo
