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8090 [49]
2 years ago
12

A sample of water with a mass of 587.00 kg is heated with 87 kJ of energy to a temperature of 518.4 K. The specific heat of wate

r is 1 J-1 kg K-1. What is the initial temperature of the water?

Chemistry
2 answers:
Tems11 [23]2 years ago
7 0

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

q = mCΔT

ΔT = q/mC = 87 kJ/[587.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2547.8 kJ/K)

= 0.035 K

Ti = Tf - ΔT = 518.4 K - 0.0354 K = 518.365 K


uysha [10]2 years ago
7 0

Answer:

370.2 K

Explanation:

The explanation is in the attached image

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How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k
Ann [662]
Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0
1 year ago
Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

Moles\ of\ methanol = 3.995\ mol

Given, Mass of water = 108 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{108\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.995\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}

Mole\ fraction\ of\ methanol=\frac{3.995}{3.995+5.995}=0.39989

<u>Mole fraction of methanol will be closest to 4.</u>

5 0
1 year ago
6. Un volumen de 1.0 mL de agua de mar contiene casi 4 x 10-12 g de Au. El volumen total de agua en los océanos es de 1.5 x 1021
Aleks [24]

Answer:

The total amount of Au is $ 2.0\times10^{24}

Explanation:

Given that,

Mass of 1.0 ml of Au m=4\times10^{-2}\ g

Total volume of water in oceans V=1.5\times10^{21}\ L

We need to calculate the volume in ml

Using given volume

V=1.5\times10^{21}\times1000\ mL

V=1.5\times10^{24}\ mL

We need to calculate the total mass of Au  

Using given data

1\ ml\ volume = 4\times10^{-2}\ g

1.5\times10^{24}\ ml=4\times10^{-2}\times1.5\times10^{24}

So, The total mass of Au is 6\times10^{22}\ g

The mass will be in ounce,

Mass=0.035274\times6\times10^{22}

Mass=2.12\times10^{21}\ ounce

The total amount of the Au Will be

Total\ amount=2.12\times10^{21}\times948

Total\ amount=2.0\times10^{24}

Hence, The total amount of Au is $ 2.0\times10^{24}

3 0
1 year ago
Describe the difference in structure between beH2 and CaH2
katen-ka-za [31]
Hello!

BeH₂ is a linear molecule, while CaH₂ is an angular molecule.

The difference between these two molecules is given by the number of electrons they have. Be is in the 2nd period of the Periodic table, and the ion Be⁺² doesn't have any free electron pairs when bonding to H. Ca is in the 4 period of the periodic table, meaning that it has more electrons, and the ion Ca⁺² has two free electron pairs when bonding to H that makes the molecule angular by pushing the bonds at an angle by sterical hindrance.  

Have a nice day!
7 0
2 years ago
Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex
zalisa [80]

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its  Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids  and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base  @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

4 0
2 years ago
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