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3 years ago

Which has greater precision, the 10 ml graduated cylinder or the 50 ml graduated cylinder?

1 answer:

7 0

The cylinder with the greater precision is the 10 ml graduated cylinder.
Precision refers to the closeness of a measurement to the true value.
A 10 ml graduated cylinder will have greater precision because it has marks for every 2 ml, this will allow you to read your measurements to two significant figures.
The 50 ml graduated cylinder on the other hand has marks for every 1 ml, this graduation will only allow the user to record measurement up to one significant figure. And you know, the more the significant figures, the more precise the measurement are.

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A protein is Select one: a. a saturated ester of glycerol. b. a polysaccharide. c. an aromatic hydrocarbon. d. a polymer of amin

Rzqust [24]

Answer:

Explanation:

The building blocks or monomers of proteins are called amino acids. There are 20 different kinds of amino acids. The amino acids form long chains that are proteins. Therefore, proteins are polymers of amino acids.

Proteins are very important for the body because they have many different roles. They provide structure for tissues, act as enzymes, hormones and antibodies, aid in transportation and fluid regulation.

Therefore, the correct answer is D. a polymer of amino acids.

7 0

2 years ago

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

d1i1m1o1n [39]

The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision

Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
For car:mass=2.3×10^3, velocity=-11
For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
24200 = -12100 + 2.3×10^3V
2.3×10^3V = 36300
V = 15.7 m/s

6 0

2 years ago

Read 2 more answers

The energies of the bonds broken in a certain reaction are greater than the energies of the bonds formed. Which one of the follo

frosja888 [35]

Answer:

a. The reaction is endothermic.

Explanation:

The heat involved in a chemical reaction is given by the enthalpy change (ΔH), which is equal to the balance between the chemical bonds that are broken (release energy) and the chemical bonds that are formed (need energy):

ΔH ≅ bonds broken - bonds formed

If broken bonds > bonds formed ⇒ ΔH > 0 ⇒ endothermic reaction

Therefore, the reaction is endothermic (it requires energy).

4 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Describe how naturally<br> acidic rainwater can affect a<br> mountain of limestone.

Hitman42 [59]

The naturally acidic rainwater will erode the limestone mountin over time

6 0

2 years ago