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Varvara68 [4.7K]

2 years ago

15

The highly reactive elements in group 7A are known for forming salts. What are they called? transition metals metalloids halogen

s noble gases

1 answer:

liberstina [14]2 years ago

4 0

The group 7A elements are called Halogens.

Please mark as brainliest if this helped! :)

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Who made the first periodic table

ale4655 [162]

The other dude is rude wrong

The correct answer is Dmitri Mendeleev

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2 years ago

Suppose you are working in a chemistry lab and trying to make a solution of a substance called Sodium Chloride (which is a solid

Alenkasestr [34]

Answer:

a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer

Explanation:

The lab apparatus that would be needed to prepare a solution of sodium chloride would be <em>a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer.</em>

The weighing balance would be used to weigh out the required amount of sodium chloride. The beaker or flask would be placed on the weighing balance and its weight zeroed. The spatula would then be sued to take out the sodium chloride from its container into the beaker till the required amount is reached. The measuring cylinder would then be used to measure out the required volume of water which would be added to the salt in the beaker. The stirrer would then be used to stir the mixture in order for the salt to dissolve.

7 0

2 years ago

What does the oxidizing agent do in a redox reaction?

WITCHER [35]

I think the answer is c not completely sure

6 0

2 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

2 years ago

A reaction is started with both reactants and products and proceeds in the reverse direction to reach equilibrium. Which of the

nekit [7.7K]

Answer:

Option A

Explanation:

A) Yes. The reaction reaches equilibrium when the rate of reaction of the reverse reaction is equal to the rate of the forward reaction , then the only cause for the reverse reaction to be favoured is that the initial rate of the reverse was greater than the forward one.

B) No. The rate constant of the reverse reaction can be greater than the forward one but the rate also depends on concentrations, thus a reverse reaction with greater rate constant can result in the net reaction proceeding in the forward reaction, the reverse reaction or be at equilibrium depending on the concentrations or reactants and products

C) No. A lower activation energy means a higher rate constant , but a higher rate constant does not mean that the net reaction will proceed to the reactants ( see point B)

D) No. The energy changes determine conditions under thermodynamic equilibrium and therefore the net direction of the reaction will depend on the temperature and concentrations of reactants and products with respect to the equilibrium conditions.

6 0

2 years ago