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Yuki888 [10]
3 years ago
11

Identifying map types

Chemistry
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

Political Map. A political map shows the state and national boundaries of a place. ...

Explanation:

You might be interested in
Will iron fillings and lead oxide powder react
Andru [333]

Answer:

Yes they woll react.

Explanation:

Ferous oxide and zinc will be product.

8 0
3 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
3 years ago
Find the percent composition of C7H5N3O6
Oksana_A [137]

Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.

First, let's find the total mass by using the masses of the elements given on the periodic table.

7 x 12.011 (mass of Carbon) = 84.077

5 x 1.008 (mass of Hydrogen) = 5.04

3 x 14.007 (mass of Nitrogen) = 42.021

6 x 15.999 (mass of Oxygen) = 95.994

Add all of those pieces together.

84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.

Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %

Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %

Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %

Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %

You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.

5 0
3 years ago
There are two different compounds of sulfur and fluorine.
olga nikolaevna [1]
What's your question......................
5 0
3 years ago
An 11.75 g sample of a common hydrate of cobalt(ii) chloride is heated. after heating, 9.25 g of anhydrous cobalt chloride remai
Irina-Kira [14]
Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt 
                                       = 11.75 g - 9.25 g = 2.50 g
number of water moles  = 2.50 g / 18 g/mol = 0.139 mol 
number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol 
ratio of water moles to CoCl₂ moles - 0.139 mol / 0.0712 mol = 1.95 
rounded off 2 moles of water for every 1 mol of CoCl₂
formula - CoCl₂.2H₂O
name - Cobalt(II) chloride dihydrate
3 0
3 years ago
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