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neonofarm [45]
2 years ago
15

4. The reaction of silver nitrate and potassium bromide yields silver bromide and potassium nitrate. If

Chemistry
1 answer:
Hatshy [7]2 years ago
4 0

Answer:

1.) AgNO₃

2.) 0.563 moles AgBr

Explanation:

The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).

AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)

Molarity (M) = moles / liters

100 mL = 1 L

AgNO₃

45.0 mL / 100 = 45.0 L

1.25 M = ? moles / 0.450 L

? moles = 0.563 moles

KBr

75.0 mL / 100 = 0.750 L

0.800 M = ? moles / 0.750 L

? moles = 0.600 moles

In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.

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Vera_Pavlovna [14]

Beryllium is a chemical element with symbol Be and atomic number 4. It is a relatively rare element in the universe, usually occurring as a product of the spallation of larger atomic nuclei that have collided with cosmic rays.

8 0
3 years ago
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0
3 years ago
Use the following scenario to answer the question: A cell has an antiport protein on its apical surface. The cell is placed in a
balu736 [363]

Answer:

The correct answer is "Secondary active transport".

Explanation:

Secondary active transport is a form of across the membrane transport that involves a transporter protein catalyzing the movement of an ion down its electrochemical gradient to allow the movement of another molecule or ion uphill to its concentration/electrochemical gradient. In this example, the transporter protein (antiporter), move 3 Na⁺ into the cell in exchange for one Ca⁺⁺ leaving the cell. The 3 Na⁺ are the ions moved down its electrochemical gradient and the one Ca⁺⁺ is the ion moved uphill its electrochemical gradient, because Na+ and Ca⁺⁺are more concentrated in the solution than inside the cell. Therefore, this scenario is an example of secondary active transport.

6 0
3 years ago
1 point
inysia [295]

Answer:

The fungus has grown larger

Explanation:

Because where the orange is in the fridge and even normally you out oranges on the counter or in a bowl, where it's in the fridge it got old faster.

8 0
3 years ago
Based on periodic properties, choose the more metallic element from each of the following pairs.
dimulka [17.4K]

Answer:

Sr is the more metallic element

Bi is the more metallic element

O is the more metallic element

As is the more metallic element

Explanation:

One thing should be clear; metallic character increases down the group but decreases across the period.

Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.

This is the basis upon which decisions were made about the metallic character of each of the elements listed above.

7 0
3 years ago
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