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3 years ago

Temperature Dependence of the pH of pure Water

Physics

1 answer:

Mamont248 [21]3 years ago

8 0

Answer:

At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

Explanation:

nonde

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A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read

timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0

3 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

$V = 21\% * 0.66L$

$V = 0.1386L$

Our values are given as

P = 1atm

T=293K $R = 0.083145kJ*mol^{-1}K^{-1}$

Using the equation to find n, we have:

$PV=nRT$

$n = \frac{PV}{RT}$

$n = \frac{(1)(0.1386)}{(0.0821)(293)}$

$n = 5.761*10^{-3}mol$

Number of molecules would be found through Avogadro number, then

$\#Molecules = 5.761*10^{-3}*6.022*10^{23}$

$\#Molecules = 3.469*10^{21} molecules$

7 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

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2 years ago

FREE BRAINIEST IF YOU ANSWER THIS

Inessa [10]

The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

w: Watts

t: Time

E: Energy required

(Watts times time is equal to the energy required)

<u>Input our values:</u>

400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

<u>Divide both sides by 400 to isolate t:</u>

<u /> $\frac{400t}{400}$ = $\frac{3000}{400}$

t = 7.5 (s)

<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

<u></u>

If you have any questions on how I got to the answer, just ask!

- breezyツ

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3 years ago

A 12.0 cm object is 9.0 cm from a convex mirror that has a focal length of -4.5 cm. What is the distance of the image from the m

Rasek [7]

Answer:

- 3 cm

Explanation:

From the mirror formula;

1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.

1/-4.5 = 1/9 + 1/v

1/v = -1/4.5 - 1/9

= -1/3

Therefore;

v = -3 cm

Hence;

Image distance is - 3cm

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2 years ago