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3 years ago

11

sound travels at 331m/s. How long would it take for sound to travel across lake tahoe, which is 21.6 miles long

1 answer:

vesna_86 [32]3 years ago

7 0

Answer:

It would take sound <em><u>104.998 seconds</u></em> to travel across lake tahoe.

Explanation:

First let's convert the miles to meters, so, 21.6 miles = 34754.4 meters

Now, we will divide this number in meters by 331. So, 34754.4 / 331 = 104.998

And what will the unit be? Seconds will be the unit for time here so 104.998 seconds.

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Read 2 more answers

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

3 years ago

Please solve this for 13 points

Anit [1.1K]

Answer:

sodium

Explanation:

everybody already known its sodium

6 0

3 years ago