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VLD [36.1K]
3 years ago
13

A student practicing for a track meet ran 300 meters in 30 seconds. what was her average speed?​

Chemistry
1 answer:
ololo11 [35]3 years ago
7 0

Answer:

<h2>10 m/s</h2>

Explanation:

Her average speed can be found by using the formula

v =  \frac{d}{t}  \\

d is the distance

t is the time taken

From the question we have

v =  \frac{300}{30}  = 10 \\

We have the final answer as

<h3>10 m/s</h3>

Hope this helps you

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Help me plz I have an 87 in this class
iVinArrow [24]

87 is a really good grade! anyway the answer is homogeneous mixture :)

7 0
3 years ago
Read 2 more answers
Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the
Alecsey [184]

Explanation:

The chemical equation is as follows.

      \frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)

And, the given enthalpy is as follows.

    \frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g);  \Delta H = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   \Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}

    102.5 = [(\frac{1}{2})x + 498] - [(2)(243)]

    102.5 = (\frac{1}{2})x + 498 - 486

     102.5 - 12 = \frac{x}{2}

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       \Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})

              x = [(\frac{1}{2})181 + (\frac{1}{2})498] - 243

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0
3 years ago
 
kirill115 [55]

Answer:c c c c c c c c c c c c c c c c c c c c c c c c c c cc c c c c c c c c

cc

Explanation:

6 0
3 years ago
How many atoms are present in 48.60 g of Mg
Ganezh [65]
1.204 x 10 ( 24 ) atoms of magnesium are present.
In order to go from mass of magnesium there have to do 2 things 

1. convert mass of Mg to moles of Mg using the molar mass of Mg as converted to a factor 

2. convert moles of Mg to atoms of Mg using the Avogadro's number ( 6.02 x 10  (23) as a conversion factor 
 
Hope my it helped a little :)  <span />
3 0
4 years ago
3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent wou
juin [17]
<span>2Al + 3Br2 -------------> 2AlBr3

</span>3 g Al = 0.11 mol Al. 

<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>

<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>

<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. 
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>


</span>
8 0
4 years ago
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