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g100num [7]
3 years ago
7

One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the

reservoir (in J/K)?
Chemistry
1 answer:
Sidana [21]3 years ago
3 0

Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

Change in heat (ΔH) = 150 joules

Temperature (T) = 150 K

Find:

ΔS surrounding (entropy change of the reservoir)

Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

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How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

6 0
1 year ago
How did you use QPOE?​
Jlenok [28]

QPOE Files

The x-ray data are stored in QPOE files (Quick Position-Ordered Events, *.qp) rather than image arrays. These are lists of photons identified by several quantities, including the position on the detector, pulse height, and arrival time. Note that, unlike IRAF images, QPOE files have no associated header file, and are always stored in the current directory, unless explicitly specified otherwise. Non-PROS IRAF tasks can also access QPOE data files in place of image arrays.

8 0
2 years ago
6.201cm+7.4cm+0.68 cm+12.0cm
zimovet [89]
If you want the answer is centimeters is it going to be:
26.281 cm
If it make it easier for you to solve add 6.201 and 7.4 which will equal 13.601. Then add .68 to 13.601 which equals 14.281. Last add 12 to 14.281 which equals 26.281.

Hope this helps.

8 0
2 years ago
What orbital is represented by the transition metals in period four
andreev551 [17]

The name transition metal refers to the position in the periodic table of elements. The transition elements represent the successive addition of electrons to the d atomic orbitals of the atoms. In this way, the transition metals represent the transition between group 2 (2A) elements and group 13 (3A) elements.

8 0
3 years ago
The radioactive decay of a certain sample produced 846 disintegrations per minute. exactly 3.00 days later, the rate of decay wa
Margarita [4]

Answer:

\boxed{\text{1.81 da}}

Explanation:

1. Calculate the decay constant

The integrated rate law for radioactive decay is 1

\ln\dfrac{A_{0}}{A_{t}} = kt

where

A₀ and A_t are the counts at t = 0 and t

k is the radioactive decay constant

\ln \dfrac{846}{269} = k \times 3.00\\\\\ln3.145 = 3.00k\\1.146 = 3.00k\\\\k =\dfrac{1.146}{3}\\\\k = \text{0.382 /da}\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{0.382} = \text{1.81 da}

The half-life for decay is \boxed{\textbf{1.81 da}}.

3 0
3 years ago
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