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jeyben [28]
3 years ago
8

What is the outcome of all chemical changes when two substances are combined?​

Chemistry
1 answer:
solmaris [256]3 years ago
7 0

Answer:

This is called a chemical reaction. gas may form.heat may be produced and the color may change

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Which of the following is an incorrect statement?
raketka [301]
Sound is a form of energy is incorrect

Sound is caused by the movement or vibration of matter

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Is barium peroxide Ba2O2 or BaO2?
vredina [299]

Answer:options?

Explanation:

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3 years ago
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consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

\frac{44}{56}

= 0.785 g of CO2

So,

<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

<u>Needed = 11.31 g</u>

<u>Available CO2  = 13.8 g</u><u> </u>(given)

So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g  of CaO produce =

\frac{100}{56}

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

\frac{Actual\ yield}{Theoritical\ yield}\times 100

\frac{19.4}{25.71}\times 100

= 75.44 %

6 0
3 years ago
How does current change.
borishaifa [10]

Answer:

if the voltage is increased, the current will increase provided the resistance of the circuit does not change.

Explanation:

5 0
2 years ago
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A chromium oxide compound contains 104.0 grams of chromium and 48.0 grams of oxygen. What is the most likely empirical formula o
TEA [102]

Answer: Empirical formula of this compound is Cr_2O_3

Explanation:

Mass of Cr= 104.0 g

Mass of O = 48.0 g

Step 1 : convert given masses into moles.

Moles of Cr =\frac{\text{ given mass of Cr}}{\text{ molar mass of Cr}}= \frac{104.0g}{52g/mole}=2moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.0g}{16g/mole}=3moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cr = \frac{2}{2}=1

For O =\frac{3}{2}=1.5

Converting into simple whole number ratios by multiplying by 2

The ratio of Cr : O= 2: 3

Hence the empirical formula is Cr_2O_3

The most likely empirical formula of this compound is Cr_2O_3

4 0
3 years ago
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