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4 years ago

Convert 3.00 x 10^5 km/sec to miles/hr. (1 mile = 1.609 km)

Chemistry

1 answer:

SVETLANKA909090 [29]4 years ago

3 0

Answer:

$3\times 10^5\ \dfrac{km}{s}=6.71\times 10^8\ \text{miles per hour}$

Explanation:

In this problem, we need to convert $3\times 10^5\ km/s$ to miles per hour

We know that,

1 mile = 1.609 km

1 hour = 3600 seconds

$3\times 10^5\ \dfrac{km}{s}=3\times 10^5\dfrac{(\dfrac{1}{1.609}\ \text{miles})}{\dfrac{1}{3600}\ \text{hour}}\\\\=6.71\times 10^8\ \text{miles per hour}$

Hence, $3\times 10^5\ \dfrac{km}{s}=6.71\times 10^8\ \text{miles per hour}$

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Suppose water did not form hydrogen bonds.

Vilka [71]

-70°C

Sink

little

hydrogen bonding

Explanation:

Completing the statements:

Water's boiling point would have been close to -70°C. Ice would sink in water. Water would release little heat to warm land during the winter. Ice is less dense than water because of the hydrogen bonding that forms a hexagonal structure in water.

The unique property of water is as a result of its hydrogen bonding. Water is a polar covalent compound. Like most covalent compound, water would have naturally had a very low boiling point.

The intermolecular forces all hydrogen bonding gives water its unique nature.

Hydrogen bond is formed by an attraction between hydrogen one water water molecule and more electronegative atom on another molecule usually oxygen, nitrogen and fluorine.

They form very strong intermolecular interaction responsible for the behavior of water.

The higher specific heat capacity of water is due to this bond. It absorbs a lot of heat and does not release them on time. This causes water release heat during winter.

Water has a hexagonal shape or structure linking each molecules.

learn more;

Hydrogen bonding brainly.com/question/10602513

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

Consider the following types of electromagnetic radiation:

liq [111]

Explanation:

Electromagnetic wave Wavelength

(1) Microwave = 1 m to 1 mm = $10^9 nm$ to $10^6 nm$

(2) Ultraviolet = 10 nm to 400 nm

(3) Radio waves = 1 mm to 100 km = $10^6 nm$ to $10^{14}nm$

(4) Infrared = 700 nm to 1 mm

(5) X-ray = 0.01 nm to 10 nm

(6) Visible = 400 nm t0 700 nm

a) In order of increasing wavelength:

: 5 < 2 < 6 < 4 < 1 < 3

b) Frequency of the electromagnetic wave given as:

$\nu=\frac{c}{\lambda }$

$\nu$ = frequency

$\lambda$ = Wavelength

c = speed of light

$\nu \propto \frac{1}{\lambda }$

So, the increasing order of frequency:

: 3 < 1 < 4 < 6 < 2 < 5

c) Energy(E) of the electromagnetic wave is given by Planck's equation :

$E=h\nu$

$E\propto \nu$

So, the increasing order of energy:

: 3 < 1 < 4 < 6 < 2 < 5

6 0

3 years ago

Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g

Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒ $Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}$

On substituting the values, we get

⇒ $0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}$

⇒ $Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}$

⇒ $=1.5\times 10^{-3}$

hence,

⇒ $Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}$

On substituting the values, we get

⇒ $1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}$

⇒ $Mass \ of \ acid=1.5\times 10^{-3}\times 400$

⇒ $=0.60 \ g$

8 0

3 years ago

why do we heat the test tube in a glass beaker during a chemistry experiment and not directly over a flame?

bearhunter [10]

It is not directly over a flame because it depends on the substance you might not want to heat it too much.you never know what could happen

7 0

4 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago