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1 year ago

Which is the correctly balanced chemical equation for the reaction of KOH and H₂SO4?

Chemistry

2 answers:

mylen [45]1 year ago

8 0

Answer:

2KOH + H2SO4 > K2SO4 + 2H2O

Explanation:

firstly the order speaks, you can't have something like OKOH so the other two is eliminated. Then check if the remaining reactions balance

ludmilkaskok [199]1 year ago

8 0

KOH + H2SO4 > K2SO4 +H2O

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A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem

BartSMP [9]

Answer:

$Q_{metal} = -6799\,J$

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

$Q_{metal} = - Q_{w}$

$Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})$

$Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)$

$Q_{metal} = -6799\,J$

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3 years ago

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PLZZZZZ help ASAP!!!!!!! THANK YOU. :) i need all 3 and thank you again!

Allisa [31]

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8 0

2 years ago

Read 2 more answers

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Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

3 0

2 years ago

The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi

Ede4ka [16]

Answer:

$\Delta _fH_{C_3H_8}=-102.7kJ/mol$

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

$\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}$

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

$\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol$

Best regards.

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2 years ago

What would happen if an alkali metal was combined with a halogen?explain

RideAnS [48]

All alkail metals react with halogens vigorously to produce salt. The alkali metal would loose an electron to form and ion with the halogen

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3 years ago