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1 year ago

15

You are riding a bicycle home from school. You average 5 meters per second and you are headed south. What motion can you determi

ne from this data? A Acceleration B Distance C Speed D Velocity

1 answer:

aleksklad [387]1 year ago

4 0

Answer : The correct option is, (D) Velocity

Explanation :

Velocity : It measures the speed of an object in a particular direction. It is a vector quantity. It depends on the magnitude and the direction. The S.I unit of velocity is, $m/s$

Distance : It measures the length between the two objects. The S.I unit of distance is, meter.

Speed : It is defined as the distance traveled per unit of time. It does not have a direction. The S.I unit of speed is, $m/s$

Acceleration : It is defined as the rate of change of velocity of an object with respect to time. The S.I unit of acceleration is, $m/s^2$

Hence, the motion we can determine from the given data is, velocity.

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A 50-gram sample has a half-life of 12 days. How much material will remain after 12 days?

Elis [28]

<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life. Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>

8 0

1 year ago

Read 2 more answers

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

2 years ago

Compute the end-to-end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule with a number-

MArishka [77]

Answer:

5156 Â

Explanation:

Chains of monomers that are being linked together constitute what is called polymers. in this question; we are to compute the end to end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule.

Using the following approach:

The first step is to determine the repeating units in polymer

The number of repeating unit = average molecular weight/ atomic weight of PVC

Given that:

average molecular weight = 256,131 g/mol

atomic weight of PVC = 62.498 g

Then;

The number of repeating unit = 256,131 g/mol / 62.498 g = 4098.227143 mol

Now the distance is calculated by using the formula:

d = (C-C) × sin 109.5/2

C-C bond length = 1.54 angstroms

tetrahedral bond angle = 109.5°

Then d = (1.54) × sin 109.5/2

d = 1.258

Thus ; the end to end separation is :

4098.227143 × 1.258 = 5155.57 Â

The answer is 5156 Â (since no decimal should be included )

6 0

1 year ago

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.

Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

= 0.093155 M

To 4 significant figure = 0.0932 M

3 0

2 years ago

Read 2 more answers

Which liquid materials have strong odor and weak odor?

densk [106]

<span>Odor refers to the fragrance caused by one or more volatilized chemical compounds. It can be strong or weak. </span>
Strong odor have: Sodium Hypochlorite, <span>Muriatic Acid, Sulphuric Acid, Ammunlom Sulfide. </span> Butyl ,Butyric Acid, Pyridine

<span>Weak Odors have </span><span>Spray Glue, Dry Erase Markers, Paint cleaners.</span><span> Water.</span>

5 0

1 year ago