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3 years ago

Write the formula for calcium chloride. Explain how you determined this formula

1 answer:

7 0

Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is CaCl2

Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.

To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.

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Given the unbalanced equation:

maw [93]

Answer:

Option (2) 2

Explanation:

NO3- + 4H+ + Pb → Pb2+ + NO2 + 2H2O

The equation above can be balance as follow:

There are 3 atoms of the left side and a total of 4 atoms on the right side. It can be balance by putting 2 in front NO3- and 2 in front of NO2 as shown below:

2NO3- + 4H+ + Pb → Pb2+ + 2NO2 + 2H2O

Now the equation is balanced.

The coefficient of NO2 is 2

8 0

2 years ago

What chemical property must argon have to make it a suitable choice for light bulbs?

wlad13 [49]

Argon is a suitable choice for light bulbs because it's inert. Compared to a reactive gas like oxygen, the metal filimant would burn up in a reactive enviroment, which is why a noble gas is used.

8 0

3 years ago

In the reaction FeCl2 + 2NaOH ->Fe(OH)2(s) + 2NaCl, if 6 moles of FeCl2 are added to 6 moles of Na0H, how many moles of NaOH

pogonyaev

Answer : The correct option is, (B) 6 mole

Explanation :

Given moles of $FeCl_2$ = 6 moles

Given moles of $NaOH$ = 6 moles

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$FeCl_2+2NaOH\rightarrow Fe(OH)_2+2NaCl$

From the given balanced reaction, we conclude that

As, 1 moles of $FeCl_2$ react with 2 moles of $NaOH$

So, 6 moles of $FeCl_2$ react with $\frac{2}{1}\times 6=12$ moles of $NaOH$

From this we conclude that, $FeCl_2$ is an excess reagent and $NaOH$ is a limiting reagent because the given moles are less than the required moles and it limits the formation of product.

Thus, the number of moles of NaOH used up in the reaction = Required moles of NaOH - Given moles of NaOH

The number of moles of NaOH used up in the reaction = 12 - 6 = 6 moles

Therefore, the number of moles of NaOH used up in the reaction will be, 60 moles

4 0

2 years ago

Read 2 more answers

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

2 years ago

Based on these data, what is the average atomic mass of element b? 10.01 10.51 10.81 11.01

Lelechka [254]

The right answer is C. 10.81

5 0

2 years ago

Read 2 more answers