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2 years ago

H2SO4 + Ca(OH)2 --> CaSO4 + H20 Balancing?

Chemistry

1 answer:

slavikrds [6]2 years ago

4 0

Answer:

This is an acid-base reaction (neutralization): H 2SO 4 is an acid, Ca(OH) 2 is a base.

...

1 H2SO4 + Ca(OH)2 → H2O + CaSO4

2 H2SO4 + Ca(OH)2 → CaSO4 + HOH

3 H2SO4 + Ca(OH)2 → H2O + CaOH2SO4

4 H2SO4 + Ca(OH)2 → H2O + Ca(HSO4)2

5 H2SO4 + Ca(OH)2 → H2O + (CaOH)2SO4

Explanation:

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How many grams of Cl are in 585 g of CaCl2

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I'm in the car now so can't lookup periodic table. But you find molecular weight of: 1 Ca + 2 Cl
Then that

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2 years ago

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enyata [817]

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3 years ago

Determine the molecular weight of H20. o 18 g O 18.0 g O 18.01 g O 18.016 g

Sonbull [250]

Answer:

The correct answer is D

18.016 g

Explanation:

Molecular Weight : It is the sum of atomic weights of each atoms present in the compound.

The molecular weught is measured in atomic mass.unit( amu) or simply"u"

This is calculated by using :

Molecular weight = number of atom x atomic mass of the atom

For H2O

Number of H atoms = 2

Number of O atom = 1

Molecular weight of H2O = 2(mass of H atom) + 1(mass of O atom)

Molecular weight = 2(1.00784) + 15.999

= 18.01558 u

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8 0

2 years ago

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 8.00 atm, PB = 5.40 atm, PC = 6.90

Yuliya22 [10]

Answer: The standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

Explanation:

For the given chemical equation:

$A(g)+2B(g)\rightleftharpoons C(g)+D(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}$

We are given:

$p_A=8.00atm\\p_B=5.40atm\\p_C=6.90atm\\p_D=5.90atm$

Putting values in above equation, we get:

$K_p=\frac{6.90\times 5.90}{8.00\times (5.40)^2}\\\\K_p=0.174$

To calculate the standard Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G^o$ = standard Gibbs free energy

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 0.174

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.174)\\\\\Delta G^o=4332.5J/mol=4.33kJ/mol$

Hence, the standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

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3 years ago

When a 0.764 g almond is burned, it heats 225 mL of water from 14.8oC to 18.4oC. Calculate the heat of combustion of the almond

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2 years ago