How many fewer H+ ions are there in a solution at a pH = 11 than in a solution at a pH = 6?

In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g

Veseljchak [2.6K]

Answer:

The mass of the ice added = 16.71 g

Explanation:

The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

Heat gained by the ice = Heat lost by the 326 g of water.

Let the mass of ice be m

The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

m = unknown mass of ice

C = Specific Heat capacity of ice = 2.108 J/g°C

ΔT = change in temperature = 0 - (-5.5) = 5.5°C

Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

Heat used by the ice to melt at 0°C = mL

m = unknown mass of ice

L = Latent Heat of fusion of ice to water = 334 J/g

Heat used by the ice to melt at 0°C = m×334 = (334m) J

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

m = unknown mass of water (which was ice)

C = Specific Heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 15 - 0 = 15°C

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

Heat lost by the water in the calorimeter cup = MCΔT

M = mass of water in the calorimeter cup = 326 g

C = specific heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 20 - 15 = 5°C

Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

Heat gained by the ice = Heat lost by the 326 g of water.

408.384m = 6,823.18

m = (6,823.18/408.384)

m = 16.71 g

Hope this Helps!!!

7 0

3 years ago

An object is placed 250 cm in front of a concave circular mirror, and the image of the object also appears at 250 cm in front of

Scrat [10]

Answer:

Radius of curvature of the mirror = 250 cm

Explanation:

Given:

Object distance from mirror = 250 cm (u=-250)

Object distance appears in mirror = 250 cm (v=-250)

Find:

Radius of curvature of the mirror

Computation:

Using mirror formula

1/f = 1/v + 1/u

1/f = 1/(-250) + 1/(-250)

f = (-250/2)

f = -125 cm or 125 cm

Radius of curvature of the mirror = 2(f)

Radius of curvature of the mirror = 2(125)

Radius of curvature of the mirror = 250 cm

3 0

3 years ago

A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H

SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

$v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s$

Time taken by the ball to reach the highest point is 2.44 seconds

a)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m$

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s$

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

$v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s$

The ball will hit the ground at 25.2117 m/s

8 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

Nuclear sizes are expressed in a unit named

o-na [289]

Answer:

Answer is A) Fermi

Explanation:

Fermi is the expressive unit for nuclear sizes. Fermi = 10^-15 meter.

4 0

3 years ago