What is a trial? please

The standard length of a certain golf club is 116.84 cm. You measure the club to be 116.85 cm long. You measure it again and fin

IgorC [24]

Answer:

The first measurement is 116.84 cm, and the second measurement is 116.85 cm.

Explanation:

The problem provides both the first and second measurements so all you have to do is fill in the blank with them.

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

a chemical reaction produced 2.50 moles of nitrogen gas. what volume in liters, does this gas sample occupy at STP? (show your w

Harrizon [31]

The volume of N₂ at STP=56 L

<h3>Further explanation</h3>

Given

2.5 moles of N₂

Required

The volume of the gas

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.

So for 2.5 moles gas :

$\tt 2.5\times 22.4=56~L$

5 0

3 years ago

Cloudy nights can be warmer than clear nights because clouds trap heat

tigry1 [53]

Answer:

Cloudy nights can be warmer than clear nights because clouds trap heat absorbed by Earth during daylight hours.

4 0

3 years ago

Read 2 more answers

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

7 0

3 years ago

What is a trial? please​

What is a trial? please