All categories

3 years ago

Why is it better to give the actual speed of an object rather than to simply say it moved fast or it moved slowly

1 answer:

3 0

Specificity. It’s really loose to say that something is fast, since speed can be scalarly linked and relative. I could say that both a car on the highway is fast, but so is the speed of light. The actual speed of something helps to do away with the arbitrary nature of using “fast” and “slow”; however, we’re still at step one of the person who is receiving the information is unfamiliar with the scale that the actual speed is defined in.

You might be interested in

PLZ I NEEEEEEEDDDDDD HELP

MrRa [10]

Answer:

2h+02=h20

Explanation:

2 in front of h on left side

2 in front of h on right side

3 0

3 years ago

Read 2 more answers

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

A ___________________________ is made of two or more different substances.

Dmitry [639]

Mixture/ compound
hope this helps

6 0

3 years ago

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion fo

Anna007 [38]

Answer:

NaCl: ionic, HF: hydrogen bond, HCl: dipole dipole , F2: dispersion force

Explanation:

complete question is:

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.Identify each of the following ( NaCl, HF, HCl, F2) as Ionic, H Bonding, Dipole or Dispersion.

3 0

3 years ago

Read 2 more answers

Calculate the amount of grams of water that was used when 5525J of energy was used to boil this amount of water.

djverab [1.8K]

Answer:

2.445 g

Explanation:

Step 1: Given and required data

Energy in the form of heat required to boil the water (Q): 5525 J

Latent heat of vaporization of water (∆H°vap): 2260 J/g

Mass of water (m): ?

Step 2: Calculate the mass of water

We will use the following expression.

Q = ∆H°vap × m

m = Q / ∆H°vap

m = 5525 J / (2260 J/g)

m = 2.445 g

4 0

2 years ago