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Marta_Voda [28]
2 years ago
9

Suppose now that you wanted to determine the density of a small crystal to confirm that it is silicon. From the literature, you

know that silicon has a density of 2.33 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of ( = 1.492 g/mL) and ( = 2.890 g/mL)? (Note: 1 mL = 1 Cm^3)
V(CHCl3)=

V(CHBr3)=
Chemistry
1 answer:
SSSSS [86.1K]2 years ago
3 0

Answer:

CHCL3= 11.73 mL

CHBr3= 8.268 mL

Explanation:

Let x be the mL of CHCl3 and y be the mL of CHBr3, then we have:

y+x = 20 mL

1.492x+2.89y=2.07* 20.0 mL

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. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dissolv
marysya [2.9K]

Answer:

3 acidic hydrogens per molecule of citric acid

Explanation:

In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:

n = 0.105x0.0372 = 0.0039 mol of NaOH

The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.

The molar mass of citric acid is:

C: 12g/mol x 6 = 72 g/mol

H: 1g/mol x 8 = 8g/mol

O: 16 g/mol x 7 = 112 g/mol

192 g/mol

So, 0.250g of the acid has

n = mass/molar mass

n = 0.250/192

n = 0.0013 mol.

To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.

The dissociation reaction of one molecule of  the acid will give the stoichiometry:

1 mol of acid ----------------------- x mol of acidic hydrogens

0.0013 mol --------------------------- 0.0039

For a simple direct three rule:

0.0013x = 0.0039

x = 3 acidic hydrogens per molecule of citric acid.

3 0
2 years ago
Write a chemical equation showing how this is an acid according to the arrhenius definition. express your answer as a balanced c
Alik [6]

According to Arrhenius theory, acid is a substance that releases H⁺ ions when dissolved in water.

In order to apply this theory, the substance must be soluble in water.

H₂SO₄ is highly soluble in water. It undergoes following dissociation reaction when dissolved in water.

H_{2}SO_{4}(aq)\overset{H_{2}O}{\rightarrow} 2H^{+}(aq)+ SO_{4}^{2-}(aq)

From the above equation, we can see that H₂SO₄ forms 2 H⁺ ions when dissolved in water. Therefore it behaves as an acid according to Arrhenius theory.

7 0
2 years ago
How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu
Alexxandr [17]
The  moles  of  chromium (iii)  nitrate  produced  is  calculated  as   follows

write  the  equation  for  reaction

 3  Pb(NO3)2  +  2 Cr  =  2 Cr(NO3)3  +  3  Pb

by  use  of  mole  ratio  between  Pb(NO3)2  to  Cr(NO3)3  which  is  3  :  2  the  moles  of  Cr(NO3)3  is therefore  
=  0.85  x2  /3  =  0.57   moles
5 0
2 years ago
Classify the following as a type of potential energy or kinetic energy (use the letters K or P)
Zanzabum
K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.
5 0
1 year ago
Read 2 more answers
Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base
Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

K_a = dissociation constant of an acid = 1.8\times 10^{-5}

K_b = dissociation constant of a base = ?

K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

1.8\times 10^{-5}\times K_b=1\times 10^{-14}

K_b=5.5\times 10^{-10}

Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

[OH^-]=1.3\times 10^{-5}M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1.3\times 10^{-5})

pOH=4.9

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1

Therefore, the pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

4 0
2 years ago
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