All categories

4 years ago

Rearrange the equation for the area of a rectangle (A=l x w) to solve for length, L

Chemistry

1 answer:

Mariana [72]4 years ago

4 0

Not sure if this is what you mean but l= A/w
(length equals area divided by width)

You might be interested in

Which of the following is the Arrhenius equation?<br>

natka813 [3]

Answer:

A is the answer the reason why is because I got it right when I was doing my test

3 0

3 years ago

If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer

Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

3 0

3 years ago

How many protons are in an element with an atomic number of 8 and a mass of 18?

Pachacha [2.7K]

B: 8 the amount of protons is equal to the atomic number

4 0

3 years ago

Read 2 more answers

The pen on seismograph swings freely.

Sever21 [200]

This statment is false, the pen on a seismograph does not swing freely.

7 0

3 years ago

El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si

Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

$PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)$

Lo cual hace que la expresión de equilibrio se calcule como:

$Ksp=[Pb^{2+}][Cl^-]^2$

Y que en términos de la solubilidad molar, s, se resuelve como:

$1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L$

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

$s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L$

¡Saludos!

7 0

3 years ago