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2 years ago

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Identify a cation. An atom that has gained a proton. An atom that has lost an electron. An atom that has gained an electron. An

atom that has lost a proton.

1 answer:

iren2701 [21]2 years ago

8 0

the answer would be B an atom that has lost an electron

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29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap

Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT = T2 - T1

ΔT = 161°C - 32°C

ΔT = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c = 499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0

2 years ago

How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100

Alona [7]

Answer:

Q = 143,921 J = 143.9 kJ.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

$Q=m*\Delta _{vap}H$

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

$Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ$

Regards!

3 0

2 years ago

if you find that a substance turns litmus paper to blue and contains hydroxide ions what would it most lightly be

Delicious77 [7]

Acid if I remeber correctly

5 0

3 years ago

2.50 g CuCl2 equals how many moles

Vitek1552 [10]

The molar mass of CuCl2 is 134.45 g/mol; therefore, you divide 2.5 g of CuCl2 by 134.45 g of CuCl2 leaving you with 0.019 moles

4 0

2 years ago

Read 2 more answers

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

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

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

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

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

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago