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aivan3 [116]
3 years ago
12

You just competed in a track meet and you run the 1500m race in 403 s. what was your average speed in miles per hour

Chemistry
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

8.46mph

Explanation:

Given parameters:

Distance run = 1500m

Time taken to cover distance = 403s

Unknown:

Average speed in mph = ?

Solution:

The average speed of a travel is the rate of change of the total distance with time.

Speed is a scalar quantity that specifies magnitude of the travel but no direction.

    Average speed = \frac{Total distance covered }{Time taken}

we need to covered distance to miles from m and time in seconds to hours.

       1609m = 1 mile

       1500m = \frac{1500}{1609} =  0.93miles

                       3600s = 1hr

                        403s  = \frac{403}{3600}  = 0.11hr

Average speed = \frac{0.93}{0.11}   = 8.46miles per hour

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Draw Au3N as a picture of atoms
Shkiper50 [21]

The picture of Au₃N is attached below.

The first part of the picture shows the formation of Au and N ions.

Formation of Au⁺¹ :

                            As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.

Formation of N⁻³ :

                            As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)

Formation of Au₃N:

                                Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.

7 0
3 years ago
Calculate the pH of a 0.50 M HIO. The Ka of hypoiodic acid, HIO, is 2.3x10–11.0.305.325.479.474.80
never [62]

Answer:

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = \frac{[H+][IO-]}{[HIO]} = 2.3 * 10⁻¹¹

At equilibrium:

  • [HIO] = 0.5 M - x
  • [H⁺] = x
  • [IO⁻] = x

<u>Replacing those values in the equation for Ka and solving for x:</u>

Ka=\frac{x^2}{0.5-x}=2.3*10^{-11} \\x^2=(2.3*10^{-11})(0.5-x)\\x^2=1.15*10^{-11}-2.3*10^{-11}x\\x^2+2.3*10^{-11}x-1.15*10^{-11}=0\\x=3.39*10^{-6}

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47

7 0
3 years ago
A.explain what is meant by term
mrs_skeptik [129]

Answer:

Explanation:

1. Atomic numbers:

Atomic number is the number of protons in an atom. It is one of the most diagonistic and representative number used in identifying an atom. The periodic table of element arranges elements based on this number.

No two elements have the same atomic number.

Protons are positively charged particles in an atom.

  • In a neutral atom, the atomic number is the same as the number electrons since electrical neutrality is attained when the number of protons and electrons are the same.
  • The atomic number determines the position of an atom on the periodic table and it is unique for every atom.

2.

Periodic patterns

These are trends on that can be predicted on the periodic table because they shew regularities down a group or sometimes across the period.

Some of these trends are atomic radius, electronegativity, metallicity, nuclear charge e.t.c

Properties of elements can be predicted using these patterns even before they are discovered.

b.

Physical property shared by Helium, Argon and Neon is that they are all gases. All group 8 elements are called noble or inert gases.

Chemical property of these elements: they are chemically unreactive.

These gases are very stable having complete electronic shell configuration. Every atom on the periodic table tries to attain the state of the noble gases.

C.

Noble Gases/ Inert Gases/ Group O elements

6 0
3 years ago
The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at
kherson [118]

<u>Answer:</u> The molecular formula of the compound is C_4H_{10}O_4

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol

  • <u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound 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=36.26g

Mass of H_2O=14.85g

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 36.26 g of carbon dioxide, \frac{12}{44}\times 36.26=9.89g 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 14.85 g of water, \frac{2}{18}\times 14.85=1.65g of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 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{9.89g}{12g/mole}=0.824moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles

  • <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 0.659 moles.

For Carbon = \frac{0.824}{0.659}=1.25\approx 1

For Hydrogen = \frac{1.65}{0.659}=2.5

For Oxygen = \frac{0.659}{0.659}=1

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

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

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is C_2H_5O_2

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 = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

n=\frac{133.9g/mol}{61g/mol}=2

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

C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4

Hence, the molecular formula of the compound is C_4H_{10}O_4

4 0
3 years ago
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