Standard 1.11 The blue and red parts of the atoms represent the?

Based on the reaction, identify the product.

amid [387]

The answer is Be ..:;

7 0

3 years ago

Why is a thermos considered a (mostly) closed system? Question 3 options: different types of liquids stay hot thermoses come in

cestrela7 [59]

Answer:

Different types of hot or cold items can be stored in a thermos and power cannot enter or exit the system when the thermos lid is tightly closed

Explanation:

Closed systems are those that do not interact or do not exchange energy with the environment that surrounds them, that is why internal temperatures and conditions are maintained.

The human body is an open system, that is, it would be the opposite of the thermos since we constantly exchange energy with the environment through sweating, emission of gases, urine, feces, and the ingestion of food.

Thermoses are systems specially created to maintain a medium, it will be maintained if its lid is hermetically closed to prevent heat leakage or entry in situations of cold fluids.

4 0

3 years ago

Write the two half-equations for the following redox reaction 2H+ +2Br- +H2SO4= Br2 + SO2 +2H2O

Oduvanchick [21]

Explanation:

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please report me :/

7 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

Hence, the amount of oxygen gas collected will be 12.8675 g

5 0

3 years ago

If irons temperature goes from 50c to 75c in a 20g sample, how many joules of heat were used?

Marina CMI [18]

Answer:

Option D. 230 J

Explanation:

We'll begin by calculating the temperature change of the iron. This can be obtained as follow:

Initial temperature (T₁) = 50 °C

Final temperature (T₂) = 75 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 75 – 50

ΔT = 25 °C

Thus, the temperature change of the iron is 25 °C.

Finally, we shall determine the amount of heat energy used. This can be obtained as follow:

Mass (M) = 20 g

Change in temperature (ΔT) = 25 °C

Specific heat capacity (C) = 0.46 J/gºC

Heat (Q) =?

Q = MCΔT

Q = 20 × 0.46 × 25

Q = 230 J

Thus, the amount of heat used was 230 J

3 0

3 years ago