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vovangra [49]
2 years ago
14

Is it possible to start with 5g of one chemical and 10g of another chemical and make 15g of a third chemical?

Chemistry
1 answer:
antiseptic1488 [7]2 years ago
3 0

Answer : No

Explanation: The chemical compounds are formed from their elements in a fixed ratio of masses, for example, in water (H2O), always Hydrogen and oxygen are combined in 1:8 by mass, i.e. with 1 g Hydrogen the Oxygen is 8 g, with 2 g Hydrogen the oxygen would be 16 g. So, there is a fixed ratio by masses.

If we were to react hydrogen with oxygen then it would again combine in 1:8 always. If we use 1 g Hydrogen with 10 g oxygen, all of the 1 g hydrogen would still react with 8 g oxygen and the third chemical formed water is 1+8 = 9 g and not 1+10 =11 g.

So, its generally not possible.

Also not all chemical reactions proceed to completion, i.e. there are always few reactant molecules left unreacted, for example the burning of candle is a chemical process where some of the lower part of wax is left behind. so, not always all of the reactants get converted to products.

Only if we have a special case where reactants react completely and you chose the masses of reactant exactly in the ratio they combine, then only the third chemica will have mass exactly equal to the sum of masses of all reactants taken at the start, like if we take hydogen 2g and oxygen 16 g then water is 2+16 = 18, because 2 will combine exactly with 16.

Hope the answer helps.

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Answer:

(C) The average molecular speed of ethane is equal to the average molecular speed of propanol.

Explanation:

When dealing with gases, you know that the temperature and speed are related. When held at a constant temperature, the speed is also held constant. We also know that ideal gases behave the same despite their identities.

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2 years ago
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What is the mass of a sample of a sample of iron (cp=0.44j/g•°C) when it is heated from 60°to 160°C. The heat is transferred is
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3 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

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