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3 years ago

1. Describe a procedure to separate a mixture of salt, sand, and water.

1 answer:

4 0

1. A filter was used to separate the sand from the salt water solution (The process of decanting was used if a filer was not available).
2. A Bunsen burner was used to boil away the water from the salt water solution leaving only salt.

I hope this helps. Let me know if anything is unclear.

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A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

4 0

3 years ago

Calculate the pOH of an aqueous solution of .0.073 M LiOH

Novosadov [1.4K]

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

pOH= 1.14

In summary, the pOH of the aqueous solution is 1.14

Learn more:

7 0

2 years ago

¿Cómo se escribe la fórmula molecular (orden de los iones)?

cluponka [151]

Answer:

A la izquierda el catión y a la derecha el anión.

Explanation:

¡Hola!

En este caso, y basado en las normas IUPAC para la escritura de las fórmulas moleculares, es necesario primero escribir el catión a la izquerda, seguido del anión a la derecha, tal y como se muestra en los siguientes ejemplos, recordando que el catión es el ion cargado positivamente y el anión, negativamente:

$K^+Cl^-\\\\Ag_2^+(SO_4)^{2-}$

Los cuales son cloruro de potasio y sulfato de plata respectivamente. También es necesario tener en cuenta que los metales tienden a ser cationes por su capacidad de perder electrones, mientras que los no metales a ganarlos y por ende resultar como aniones.

¡Un gusto ayudarte!

6 0

3 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

Answer: The time taken by the reaction is 84.5 seconds

Explanation:

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago

Paul has different spoons for cooking. He told Sue that he likes to stir hot food with a wooden spoon because it does not burn h

IceJOKER [234]

Answer:

Sue was not right as wood is not a heat conductor and will not allow heat to pass through the spoon thus keeping the hand safe. In a meatal spoon, heat can pass through and burn the hand as meatal is a good conductor of heat.

8 0

3 years ago