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garik1379 [7]
2 years ago
5

How do solve for density?

Chemistry
1 answer:
Pachacha [2.7K]2 years ago
7 0
Density(D) is defined as Mass(M) divided by Volume(V).

The formula for Density is:

D = M / V.

Another way to remember the formula for Density is to remember "Mass per unit of volume".

I hope this helps!
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How do you calculate mass using density and volume??
LiRa [457]
Just reorder the equation

Density= mass/volume
8 0
2 years ago
How are starch cells in plants and fat cells in animals similar and different
jeyben [28]
I can not answer your question because i am a  person who is not smart
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6 0
3 years ago
What is the density of a gas with a molar mass of 28.26 grams per mole at 3.510 ATM in 237 Kelvin​
Vikki [24]
<h3><u>Answer;</u></h3>

= 5.1 g/L

<h3><u>Explanation;</u></h3>

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Number of moles is 1 since one mole has a mass equivalent to the molar mass.

Therefore; We can find the volume and thus get the density.

<em>V = nRT/P</em>

<em>    = (1 × 0.08206 × 237)/3.510</em>

<em>     = 5.5408 L</em>

<em>Hence; Density = mass/volume </em>

<em>                           = 28.26 g/5.5408 L</em>

<em>                           = 5.1 g/L</em>

<em>The Density is 5.1 g/L or 0.005 g/cm³</em>

6 0
3 years ago
4 Laconcentracion delcado clorhiclrico comercial
Darya [45]
EJERCICIOS RESUELTOS DISOLUCIONES
1.- Se disuelven 20 = g de NaOH en 560 g de agua. Calcula a) la concentración de la disolución en % en masa b) su molalidad.
Ar(Na) 23. Ar(O)=16. Ar(H)=1.
1mol NaOH = X ;X =0,5moles. m= moles(soluto) ; m= 0,5moles =0,89m; 40 g 20 g m(kg) de disolvente 0,56 kg
2.- ¿Qué cantidad de glucosa, C6H12O6 (Mm = 180 g/mol), se necesita para preparar 100 cm3 de disolución 0,2 molar?
M = moles(soluto) ; moles C6 H12O6 = M.V = 0,2M.0,1l; moles C6 H12O6 = 0,02. V (l) de disolución
1 mol glu cos a = 0,02 moles ; X = 36 g. 180 g X
3.- Se dispone de un ácido nítrico comercial concentrado al 96,73 % en masa y densidad 1,5 g/mL. ¿Cuántos mL del ácido concentrado serán necesarios para preparar 0,2 L de disolución 1,5 M de dicho ácido? Mm (HNO3) = 63g/mol.
Primeramente calcularemos los moles de ácido puro que necesitamos:
M= moles(soluto) ; moles(HNO3)=M.V=1,5M.0,2l=0,3. V (l) de disolución
Ahora calculamos la masa en g correspondiente:
0,3moles x 63g = 18,9 g de HNO3 . 1mol
Como el ácido comercial del que disponemos no es puro, sino del 96,73 % necesitaremos pesar:
100g del ácido comercial = X ; X =19,54g ácido comercial. contienen 96,73g ácido puro 18,9g ácido puro
Como necesitamos averiguar el volumen en mL que hemos de coger, utilizamos la densidad del ácido comercial:
d(g/ml)= m(g) ; V(ml)= 19,54g =13ml. V ( m l ) 1, 5 g / m l
a)
%NaOH = m(g)NaOH .100; m( g )disolución
%NaOH = 20 .100; 580
%NaOH = 3,45. b) Primeramente calculamos los moles que son los 20 g de soluto:

4.- Calcula la masa de nitrato de hierro (II), Fe(NO3)2, que hay en 100 mL de disolución acuosa al 6 %. Densidad de la disolución 1,16 g/mL
De la densidad sabemos que los 100 ml de disolución tienen de masa 116 g. Como es al 6 %, la masa de soluto existente será:
En 100g disolución = En 116g disolución ; X = 6,96g Fe(NO3 )2 . hay 6g Fe(NO3 )2 X
5.- Indica de qué modo prepararías 1⁄2 L de disolución 0,1 M de HCl si disponemos de un HCl concentrado del 36 % y densidad 1,19 g/mL
Calculamos la masa de HCl que necesitamos. Para ello, utilizando el concepto de molaridad, averiguamos primeramente los moles de HCl que va a tener la disolución que queremos preparar:
n(HCl) = M.V = 0,1M.0,5l = 0,05moles.
C o m o M m ( H C l ) = 3 6 , 5 g / m o l . L o s 0 , 0 5 m o l e s s e r á n : 0 , 0 5 m o l e s . 3 6 , 5 g = 1, 8 3 g H C l .
1mol
Esa masa de HCl la tenemos que coger del HCl concentrado del que se dispone (36 % y densidad 1,19
g/ml.). Al no ser puro, sino del 36 % tendremos que coger más cantidad de gramos: 100g del HCl concentrado = X ; X = 5,08g HCl puro.
contienen 36g HCl puro 1,83g HCl puro
Como se trata de un líquido del que conocemos su densidad, determinamos el volumen de esos 5,08 g:
V=m; V= 5,08g =4,27mlHCldel36%. ρ 1,19g / ml
Preparación: En un matraz aforado de 1⁄2 l que contenga algo de agua destilada, se introducen 4,27 ml del HCl concentrado del 36 %, utilizando una pipeta. No absorber el ácido con la boca porque es tóxico.
Se agita con cuidado el matraz hasta que se disuelva el soluto.
Se añade agua destilada al matraz hasta alcanzar exactamente la señal de 500 ml.
6.- Se disuelven en agua 30,5 g de cloruro amónico (NH4Cl) hasta obtener 0,5 l de disolución. Sabiendo que la densidad de la misma es 1027 kg/m3, calcula:
a) La concentración de la misma en porcentaje en masa. b) La molaridad.
c) La molalidad.
d) Las fracciones molares del soluto y del disolvente.
Mm(NH4Cl)=53,5g/mol.
Primeramente 1027kg/m3 = 1,027 g/cm3. Luego la masa de 1 l de disolución será de 1027 g y la de medio litro 513,8 g. De ellos 30,5 g son de soluto (cloruro amónico) y el resto 483,3 g son de agua.

a) %masa NH4Cl = masa(g)soluto x100 = 30,5g x100 = 5,94%. masa(g)disolución 513,8g
b) M = moles soluto =
Planteamos la ecuación con los moles de manera que la suma de los que tomamos de la disolución A más los que tomamos de la disolución B sea igual a 0,5
1,5.V +0,1(1−V )=0,5; V =0,286l=286cm3. V =0,714l=714cm3. AAA B
8 0
3 years ago
Select all the true statements. When an atom gains an electron, it becomes a cation. Anions carry a positive charge. The Cl− and
REY [17]

Answer:

The statements 4 and 5 are true.

Explanation:

1. When an atom gains an electron it becomes negatively charged. This negatively charged species is called anion.

A + e⁻ →  A⁻ (anion)

Therefore, the statement 1 is false.

2.  An anion is formed when an atom gains an electron and becomes negatively charged. Therefore, an anion is a negatively charged species.

A + e⁻ →  A⁻ (anion)

Therefore, the statement 2 is false.

3. The atomic number of chlorine atom Cl is 17 and atomic number of bromine atom Br is 35.

Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.

Therefore, the number of electrons in Cl atom is 17 and the number of electrons in Br atom is 35.

When the Cl atom gains one electron it forms Cl⁻ ion and when the Br atom gains one electron it forms Br⁻ ion.

Therefore, the number of electrons in Cl⁻ ion is 17 + 1 = 18 electrons

and the number of electrons in Br⁻ ion is 35 + 1 = 36 electrons

Therefore, Cl⁻ and Br⁻ ions do not have the same number of electrons.

Therefore, the statement 3 is false.

4. When potassium atom (K) loses one electron it forms a positively charged species called potassium cation (K⁺).

K  → K⁺ + e⁻

Therefore, the statement 4 is true.

5. The atomic number of Fe atom is 26.

Since, the atomic number of an atom is equal to the number of protons present in that atom.

When the Fe atom loses two electrons to form Fe²⁺ and when the Fe atom loses three electrons to form Fe³⁺ ion, the number of protons remains the same.

Therefore, the ions Fe²⁺ and Fe³⁺ have the same number of protons.

Therefore, the statement 5 is true.

6. The atomic number of copper atom Cu is 29.

Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.

Therefore, the number of electrons in Cu atom is 29

When the Cu atom loses one electron it forms Cu⁺ ion and when the Cu atom loses two electrons it forms Cu²⁺ ion.

Cu  → Cu⁺ + e⁻                and    Cu  → Cu²⁺ + 2e⁻

Therefore, the number of electrons in Cu⁺ ion is 29 - 1 = 28 electrons

and the number of electrons in Cu²⁺ ion is 29 - 2 = 27 electrons

Therefore,  Cu⁺ ion and Cu²⁺ ion do not have the same number of electrons.

Therefore, the statement 6 is false.

3 0
3 years ago
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