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3 years ago

Metric prefixes 9th grade level

Chemistry

1 answer:

zhannawk [14.2K]3 years ago

6 0

Answer:

<h3>A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or submultiple of the unit. All metric prefixes used today are decadic. Each prefix has a unique symbol that is prepended to any unit symbol.</h3>

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. Consider the following half-reactions:

Vedmedyk [2.9K]

d. Fe(s) and Al(s)

<h3>Further explanation</h3>

In the redox reaction, it is also known

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal : d. Fe(s) and Al(s)

7 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

Pls helpits rlly hard

victus00 [196]

Answer:

5: 0.16

6: 50

Explanation:

Question 5:

We can use the equation density = mass/ volume.

We already have the mass (12g), but now we need to find the volume of the cylinder.

The equation for this is πr²h

So we know the radius is 2 and the height is 6.

π x (2)² x 6 = 24π = 75.398cm³

Now we can use the density equation above:

12/75.398 = 0.1592g/cm³ = 0.16g/cm³.

Question 6:

This time, we have to rearrange the equation density = mass/ volume to find the mass.

We know mass = density x volume.

From the question, the density is 2.5g/mL and the volume is 20mL.

Following the equation above, we do 2.5 x 20 to get 50g.

5 0

3 years ago

Chemical equation to show cdso4 dissolves in water

olga_2 [115]

no se bb si quieres salimos tu sabes

5 0

3 years ago

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

shutvik [7]

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

$AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}$

If pH is increased then concentration of $H^{+}$ increases in solution resulting removal of $CH_{3}COO^{-}$ by forming $CH_{3}COOH$ . Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more $AgCH_{3}COO$ will dissolve
If $AgNO_{3}$ is added then concentration of $Ag^{+}$ increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less $AgCH_{3}COO$ will dissolve
Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of $Ag^{+}$ . So, more $AgCH_{3}COO$ will dissolve

Hence all three statements are true

8 0

3 years ago