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3 years ago

11

Calculate the energy needed to raise the temperature of 165.0 g of water from 10.0°C to 40.0°C. The specific heat capacity for w

ater is 4.2J/g °C.

1 answer:

mr Goodwill [35]3 years ago

7 0

Answer:

20790 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 165 g.

Initial temperature (T1) 10 °C.

Final temperature (T2) = 40 °C.

Specific heat capacity (C) = 4.2 J/g °C.

Heat (Q) required =?

Next, we shall determine the change in temperature of water. This can be obtained as illustrated below:

Initial temperature (T1) 10 °C.

Final temperature (T2) = 40 °C.

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 40 – 10

ΔT = 30 °C

Finally, we shall determine the heat energy required to raise the temperature of the water as follow:

Mass (M) = 165 g.

Specific heat capacity (C) = 4.2 J/g °C.

Change in temperature (ΔT) = 30 °C

Heat (Q) required =?

Q = MCΔT

Q = 165 × 4.2 × 30

Q = 20790 J

Thus, the heat energy required to raise the temperature of the water is 20790 J

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A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0

3 years ago

Which of the following is a heterogeneous mixture? Air. Salt water. Steel. Soil.

kicyunya [14]

Answer: Heterogenous mixtures are the ones in which we can see the different components clearly. Air is considered to be homogenous, because we don't see the different noble gases' particles floating around. Salt water is also homogenous, because salt dissolves almost completely into water, becoming hard to see without the proper equipment salt particles in it. Steel is also homogenous, since the metals that compose it are mixed during the heating. The only heterogenous would be soil, because we can see different particles in it, with different sizes, shapes, colors, etc without having to use special equipment.

3 0

3 years ago

Read 2 more answers

Which functional group does the molecule below have?

makvit [3.9K]

Answer:

B

Explanation:

7 0

2 years ago

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

6 0

4 years ago

If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac

STALIN [3.7K]

<u>Answer:</u> The theoretical yield of barium sulfate is 50.9 grams

<u>Explanation:</u>

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

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

<u>For sodium sulfate:</u>

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol$

<u>For barium chloride:</u>

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

$\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol$

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

$Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl$

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.228=0.228mol$ of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = $\frac{1}{1}\times 0.228=0.228moles$ of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

$0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g$

Hence, the theoretical yield of barium sulfate is 50.9 grams

7 0

3 years ago