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Mariulka [41]
3 years ago
7

a substance that undergoes a physical change like melting is still the same substance is it true or false​

Chemistry
2 answers:
Jlenok [28]3 years ago
6 0

Answer:This is true

Explanation:

Its true because its chemical composition stays the same

Juli2301 [7.4K]3 years ago
3 0

the answer would be true

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JUST ONE MULTIPLE CHOICE QUESTION!
Butoxors [25]
I think it’s The second option sorry if I’m wrong
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3 years ago
Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
olganol [36]

 The  amount  of  Al2O3  in moles=  1.11 moles    while in  grams   = 113.22 grams


    <em><u>calculation</u></em>

     2 Al  + Fe2O3 → 2Fe  + Al2O3

    step  1: find the moles of Al  by  use of <u><em>moles= mass/molar  mass  </em></u>formula

    =  60.0/27= 2.22  moles


    Step 2: use the mole ratio to determine the  moles of Al2O3.

 The  mole ratio  of Al : Al2O3 is  2: 1 therefore the moles of Al2O3= 2.22/2=1.11  moles


Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

7 0
3 years ago
Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit
rosijanka [135]

Explanation:

The relation between K_a\&K_b is given by :

K_w=K_a\times K_b

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1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}

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1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}

K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}

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3 years ago
During normal, relaxed respiration, about 500ml of air enters and leaves the lungs with each respiratory cycle. this is called t
Marianna [84]

During normal respiration, about 500ml of air enters and leaves the lungs with each respiratory cycle. This is called the<u> tidal volume</u>.

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To find out more about tidal volume, visit:

brainly.com/question/17439101

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6 0
1 year ago
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The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is \frac{ V_{1} }{ T_{1} }=\frac{ V_{2} }{ T_{2} }. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.
6 0
3 years ago
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