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2 years ago

Please help me, thank you

Chemistry

2 answers:

lina2011 [118]2 years ago

6 0

Answer:

It's neither 3 or 4 but I'm a little leaning to 4 as a right answer

Bezzdna [24]2 years ago

5 0

Answer:

it's C

Explanation:

Yes

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How many milliliters of a 0.1000 M NaOH solution would be needed to neutralize a 1.0000-g sample of potassium hydrogen phthalate

7nadin3 [17]

Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085

Explanation:

7 0

2 years ago

For the reaction2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)ΔH° is -125 kJ/mol and ΔS° is +253 J/K ∙ mol. This reaction is _

ivanzaharov [21]

Answer:

C. spontaneous at all temperatures

Explanation:

The spontaneity of reaction is determined by the sign of the gibbs free energy.

A negative sign denotes that the reaction is spontaneous, positive sign means the reaction is not spontaneous.

From the question;

ΔS° = +253 J/K

ΔH° = -125 kJ/mol

ΔG = ΔH° - TΔS°

From the data given, the condition in which we can obtain a negative value of G, is at any value of T.

For any value of T, G would always be a negative value.

This means the correct option is option C.

3 0

2 years ago

Does adding milk impact the health benefits of the tea? Use your understanding from this unit about solutions and food chemistry

mezya [45]

Yes, it mixes it and has vitamins in the tea.

5 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago

Green plants use light from the sun to drive photosynthesis. photosynthesis is a chemical reaction in which water and carbon dio

Korolek [52]

The equation that represents the process of photosynthesis is:

6CO2+12H2O+light->C6H12O6+6O2+6H2O

Photosynthesis is the process in plants to make their food. This involves the use carbon dioxide to react with water and make sugar or glucose as the main product and oxygen as a by-product. Since we are not given the mass of CO2 in this problem, we assume that we have 1 g of CO2 available. We calculate as follows:

1 g CO2 ( 1 mol CO2 / 44.01 g CO2 ) ( 12 mol H2O / 6 mol CO2 ) ( 18.02 g / 1 mol ) = 0.82 g of H2O is needed

However, if the amount given of CO2 is not one gram, then you can simply change the starting value in the calculation and solve for the mass of water needed.

7 0

3 years ago