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2 years ago

14

A piece of unknown metal weighs 500 g. When the piece of metal absorbs 6.64 kJ of heat, its temperature increases from 25°C to 7

8°C. What's the specific heat capacity
of this metal?
A. 13 J/gºC
B. 0.25 J/gºC
C. 25 J/gºC
D.2.5 J/g °C

2 answers:

Vika [28.1K]2 years ago

4 0

6.64×1000÷(78-25)×500=0.25 g/k

borishaifa [10]2 years ago

3 0

Answer:

6.64×1000÷(78-25)×500=0.25 g/k

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Sulfurous acid, h2so3, breaks down into water (h2o) and sulfur dioxide (so2). if only one molecule of sulfurous acid was involve

devlian [24]

Explanation:

The given reaction equation will be as follows.

$H_{2}SO_{3} \rightarrow H_{2}O + SO_{2}$

Now, number of atoms on reactant side are as follows.

H = 2

S = 1

O = 3

Number of atoms on product side are as follows.

H = 2

S = 1

O = 3

Therefore, this equation is balanced since atoms on both reactant and product sides are equal.

Thus, we can conclude that there is one sulfur atom in the products.

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2 years ago

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In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are

mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

$W = 200*(1-0.704)\\W=59.2\ g$

There are 59.2 grams of water in this solution.

5 0

2 years ago

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

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2 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia

lapo4ka [179]

Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

$Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm$

4 0

2 years ago

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass

inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

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

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g$

Mass of air, $m_1$ = 8.56 g

<u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g$

Mass of helium, $m_2$ = 1.18 g

Calculating the mass difference between the two:

$\Delta m=m_1-m_2$

$\Delta m=(8.56-1.18)g=7.38g$

Hence, the mass difference between the two is 7.38 grams.

5 0

2 years ago