7.24 miles of Ag(PO4)2

Help me please for brainliest and 10 points.

Stella [2.4K]

Answer:

The red arrow from igneous to sedimentary rock

Explanation:

When rocks are weathered and eroded, they become sediment. This sediment is later compacted and cemented to form sedimentary rock. Therefore, the red arrow going from igneous rock to sedimentary rock represents weathering and erosion.

7 0

3 years ago

GUYS I NEED HELP THIS IS DUE TONIGHT !!!

kenny6666 [7]

Explanation:

To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.

Based on this premise, the number of moles of species on the reactant and product side must be the same;

Li + Br₂ → LiBr

Put a,b and c as the coefficient of each species

aLi + bBr₂ → cLiBr

balancing Li;

a = c

balancing Br;

2b = c

let a = 1;

c = 1

b = $\frac{1}{2}$

or a = 2, b = 1 , c = 2

2Li + Br₂ → 2LiBr

P + Cl₂ → PCl₃

Using the same method;

aP + bCl₂ → cPCl₃

balancing P;

a = c

balancing Cl;

2b = 3c

let a = 1;

c = 1

b = $\frac{3}{2}$

or

a = 2, b = 3, c = 2

2P + 3Cl₂ → 2PCl₃

iii,

H₂ + SO₂ → H₂S + H₂O

use coefficients a,b,c and d;

aH₂ + bSO₂ → cH₂S + dH₂O

balancing H;

2a = 2c + 2d

balancing S;

b = c

balancing O

2b = d

let b = 1,

c = 1

d = 2

a = 3

3H₂ + SO₂ → H₂S + 2H₂O

3 0

3 years ago

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

$\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}$

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

$Density=\frac{Mass}{Volume}$

Now putting all the given values in this formula, we get:

$3.52g/cm^3=\frac{284.4g}{Volume}$

Volume = $80.8cm^3$

As we know that:

$1cm^3=0.061inch^3$

So,

Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

Therefore, the volume of pure diamond is $0.493inch^3$

5 0

3 years ago

Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow

Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

$\frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)$

$\rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h$

$\frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}$

[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = $\frac{10}{\rho A_{c}}$

$A_{c} = 0.01 m^{2}$

$\frac{dh}{dt}$ + h = 1

$\frac{dh}{dt}$ = 1 - h

$\frac{dh}{1 - h}$ = dt

$\frac{ln(1 - h)}{-1}$ = t + C

Given at t = 0 and V = 0

$A \times h$ = 0

or, h = 0

-ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

C = 0

So, -ln(1 - h) = t

or, t = $ln (\frac{1}{1 - h})$ ........... (1)

(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

t = $ln (\frac{1}{1 - h})$

t = $ln (\frac{1}{1 - 0.6})$

= $ln (\frac{1}{0.4})$

= 0.916 seconds

(b) As maximum height of water level in the tank is achieved at steady state that is, t = $\infty$ .

1 - h = exp (-t)

1 - h = 0

h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

8 0

3 years ago

How do I find x mm? This is for my chemistry class.

Evgesh-ka [11]

Im guessing it might be 98.4x0.58, when you rearrange the pressure formula.

7 0

3 years ago