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ololo11 [35]
3 years ago
10

Earth is approximately 1.5x10^8 km from the sun. The speed of light is 3.0x10^8. How many minutes does it take for light to reac

h earth from the sun?
Chemistry
1 answer:
Masteriza [31]3 years ago
3 0
1.5 x 10^{8} mm = 1500 x 10^{8} m

1500 x 10^{8} / ( 3.0 x 10^{8} ) = 500 s

500/60 = 8.3 

So approximately it takes 8.5 minutes 


Hope that helps, Good luck! (:<span />
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What role does the wire play in voltaic cell?
Triss [41]
<h3><u>Answer;</u></h3>

a) It allows electrons to flow from the anode to the cathode.

<h3><u>Explanation</u>;</h3>
  • <em><u>Voltaic cell is an electrochemical cell in which a spontaneous chemical reaction produces the flow of electrons</u></em>.
  • Electrons are produced by the oxidation reaction occurring at the anode. Electrons flow through the conducting wire from the anode to the cathode. At the cathode these electrons are used to reduce copper(II) ions to copper atoms.
  • <em><u>A conducting wire or a wire play connects the two electrodes allowing electrons to flow from the anode to the cathode</u></em>.
7 0
3 years ago
Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be
Sphinxa [80]

Answer:

P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

Explanation:

The first step is:

P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}

Second step is:

P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}

Multiplying second step by 2, and adding both the steps, we get that:

P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}

Cancelling common species, we get that:

P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

6 0
3 years ago
What is the speed at which molecules or atoms move dependent on temperature and state of matter.
den301095 [7]
The hotter it gets, the faster molecules move, solid form is in low temperature, liquid in medium temperature and gas in high temperature.
3 0
4 years ago
The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from
balu736 [363]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 100 ml.</em></u>

Explanation:

 Given:

Initial pressure (P_{1}) = 600 mm of Hg

Final pressure (P_{2}) = 1200 mm of Hg

Initial volume (V_{1}) = 200 ml      

To find:

Final volume (V_{2})

We know;

According to the ideal gas equation,

    P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

 From the above mentioned equation,

        P × V = constant

\frac{P_{1} }{P_{2} } = \frac{V_{1} }{V_{2} }

Where,

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1})  represents the initial volume of the gas

(V_{2})  represents the final volume of the gas

So;

\frac{600}{1200} = \frac{V_{2} }{200}    

V_{2} = 100 ml

<u><em>Therefore the final volume of the gas is 100 ml.</em></u>                                                                                                                                                                              

5 0
4 years ago
Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

5 0
3 years ago
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