All categories

3 years ago

Consider the balanced chemical equation,

Chemistry

2 answers:

Gemiola [76]3 years ago

7 0

Answer: we started with 10 ions of Mg and 20 ions of F.

Explanation: The given balanced equation is:

$MgCl_2(aq)+2AgF(aq)\rightarrow 2AgCl(s)+MgF_2(aq)$

From this balanced equation, 2 formula units of AgCl are formed by 2 formula units of AgF and 1 formula unit of $MgCl_2$ .

From here, the ratio between AgCl and AgF is 2:2 that is 1:1 and the ratio between AgCl and $MgCl_2$ is 2:1 ,

Since the ratio between AgCl and AgF is 1:1, 20 formula units of AgCl will be formed by 20 formula units of AgF.

The ratio between AgCl and $MgCl_2$ is 2:1, so 20 formula units of AgCl will be formed by 10 formula units of $MgCl_2$ .

One formula unit of $MgCl_2$ contains one ion of Mg, so 10 formula units of $MgCl_2$ will contain 10 ions of Mg.

Similarly, 1 formula unit of AgF contains one ion of F, so 20 formula units of AgF will contain 20 ions of F.

So, we started with 10 ions of Mg and 20 ions of F to make 20 formula units of AgCl.

Montano1993 [528]3 years ago

4 0

<span>We started with twenty ions of Mg and twenty ions of F. Producing 20 units of AgCl requires twenty MgCl2 since it is the only source of magnesium. Similarly it requires twenty AgF since it is the only source of silver.</span>

You might be interested in

The decomposition of water into hydrogen gas Hy and oxygen gas Oz can be modeled by the balanced chemical equation

ASHA 777 [7]

Answer:

D. ) 2H2O - 2H2 + O2

3 0

3 years ago

In the reaction of aluminum hydroxide and caldium nitrate, how many grams of calcium hydroxide will be formed if 12.55 g of alum

Dahasolnce [82]

Answer:

17.76g

Explanation:

We need to write a balanced chemical equation for the reaction:

2Al(OH)3 + 3Ca(NO3)2 ——> 3Ca(OH)2 + 2Al(NO3)3

In the reaction above, it can be seen that 2 moles of aluminum hydroxide yielded 3 moles of calcium hydroxide. This is the theoretical viewpoint.

Now we need to know what actually happened. We need to calculate the actual number of moles of aluminum hydroxide reacted l. We can get this by dividing the mass by the molar mass.

The molar mass of aluminum hydroxide is 27+ 3( 16+1)

= 27 + 51 = 78g/mol

The number of moles is thus: 12.55/78 = 0.16 moles

Now if 2 moles of aluminum hydroxide gave 3 moles of calcium hydroxide, 0.16moles will give : (0.16*3)/2 = 0.24moles

Now we can calculate the mass of calcium hydroxide formed. The mass of calcium hydroxide formed is the number of moles multiplied by the molar mass.

The molar mass of calcium hydroxide is; 40 + 2(17) = 74g/mol

The mass is thus =74 * 0.24 = 17.76g

7 0

3 years ago

What statement does not accurately describe the following equation? C6H12O6+--->6H20+6CO2+energy A) oxygen is a react ant B)

STatiana [176]

The answer is B. The complete equation is C6H12O6 + 6O2 -->6H2O + 6CO2 + energy. So we can know that A and C and D is right. For B, the reaction release energy so it is exothermic reaction.

4 0

3 years ago

Read 2 more answers

Why does an atom of 23NA have 23 electrons

Julli [10]

It is actually something standardized more so than theoretical, however in terms of atoms in general the electrons are based on the amount of electronic shells that an atom has and the amount of electrons that atom can accommodate. In the case of sodium, it has three shells with 2 electrons on the first shell which is the maximum, 8 on the second shell which is also the max and 1 on their final shell
So simply put an ATOM of Sodium (Na) has 23 electrons because it has 23 protons which is a fact
<span />

7 0

3 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

3 years ago