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3 years ago

How does chemical change affect the composition of matter

Chemistry

1 answer:

Sergio039 [100]3 years ago

8 0

Beacuse it gets oxagin and changes ur wlecome

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How many grams are in 3.4 moles of Lead (IV) Sulfate (Pb(SO4)2)? ) Molar mass: 303.26 g/mol 131.0840

Murrr4er [49]

Answer:

1031.084g

Explanation:

n=m/M

m=n(M)

m = (3.4mol)(303.26g/mol)

1031.084g

Hope that helps

4 0

2 years ago

Read 2 more answers

Question 1

maxonik [38]

Answer:

it is 2 degrees Celcius

Explanation:

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3 0

3 years ago

Determine the freezing point depression of 2 kg of water when 2 mol salt is added to it. The kf of water is 1.86 degrees C/M. Sh

Svet_ta [14]

Answer: The freezing point depression is $1.86^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f$ = Depression in freezing point

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality = $\frac{\text {moles of solute}}{\text {mass of solvent in kg}}=\frac{2mol}{2kg}=1m$

$\Delta T_f=1.86mol/kg^0C\times 1m$

$\Delta T_f=1.86^0C$

Thus freezing point depression is $1.86^0C$

3 0

3 years ago

Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of carbon dioxide would be produced by

Bas_tet [7]

Answer:

$V_{CO_2}=9.23m^3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Now, as the stoichiometrical factors are in terms of mole but no information about neither the temperature nor the pressure is given, by means of the Avogadro's law, one could perform the stoichiometric calculations with the given volume as both the pressure and temperature remain the same, that is:

$V_{CO_2}=9.23m^3CH_4*\frac{1m^3CO_2}{1m^3CH_4} \\\\V_{CO_2}=9.23m^3$

Such 1:1 volume relationship equals the 1:1 molar relationship given in the chemical reaction in terms of their stoichiometric coefficients, therefore, the yielded volume of carbon dioxide is also 9.23m³

Best regards.

4 0

4 years ago

How many molecules of CaCL2 are equivalent to 75.9g CaCl2

disa [49]

The answer is around 4.12x10^23 molecules

3 0

3 years ago