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what is the relationship between the scale divisions marked on the graduated cylinders and the estimated uncertainty in the volu

AURORKA [14]

Answer:

The relationship between the scale divisions marked on the graduated cylinders and the estimated uncertainty in the volume measurements is that the larger the divisions the larger the uncertainty, and the smaller the divisions the smaller the uncertaintity.

Explanation:

In general, graduated cylinders are marked with divisions (lines) at equal spaces or intervals. There are divisions that show the measure with a number and a series of intermediate not numbered divisions in between two numbered lines.

The uncertainty of the instrument is determined as half the difference between two consecutive marks.

So, the larger the interval of two consecutive marks the larger the uncertainty.

The attached image, taken from brainly.com/question/13102373, shows you some marks in a graduated cylinder.

7 0

3 years ago

A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate

Dmitry [639]

Answer: The freezing point of solution is -5.11°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

$NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)$

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute (NaCl) = 7.57 g

$M_{solute}$ = Molar mass of solute (NaCl) = 58.44 g/mol

$W_{solvent}$ = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

$\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 2

$K_f$ = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

$\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC$

Hence, the freezing point of solution is -5.11°C

7 0

4 years ago

A student has four resistors. Each resistor has a resistance of 100 ohms.

ValentinkaMS [17]

Answer:

The minimum resistance is 25 ohms.

Explanation:

Resistance of each resistor is 100 ohms. When resistors are connected in parallel, the equivalent resistance is lowest. For parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....$

Here, all resistors are 100 ohms. So,

$\dfrac{1}{R}=\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}+\dfrac{1}{100}\\\\R=25\ \Omega$

So, the minimum resistance is 25 ohms.

7 0

3 years ago

21 pts. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts

Aleksandr [31]

First, calculate the amount fo Na2O that should be produced with the given amount of sodium using proper dimensional analysis and the balanced chemical reaction,

theoretical amount of Na2O = (54.1 g of Na)(1 mol Na/23 g Na)(2 mos Na2O/4 mols Na)(62 g Na2O/ 1mol Na2O)

theoretical amount of Na2O = 72.9 g of Na2O

Calculation of percent yield,

% yield = (61.8 g Na2O) / (72.9 g of Na2O) x 100%
% yield = 84.75%

Answer: 84.75%

8 0

3 years ago

What is an element in science

s2008m [1.1K]

An element is a part or aspect of something abstract, especially one that is essential or characteristic. I hope this helps you good luck :)

5 0

3 years ago

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