A living thing that cannot make its own food and must eat other living things for energy

Which of the following chemical equations is NOT correctly balanced?

forsale [732]

To see if an equation is balanced, make sure that each side of the reaction has an equal number of the types of atoms. The one that has a mismatch is A

6 0

3 years ago

The following experiment was performed: A white solid was added to a test tube containing water. The solid completely dissolved

Morgarella [4.7K]

Answer:

Potassium iodide

Explanation:

The white solid introduced into the solution is potassium bromide. This is an ionic solid thus it quickly dissolves in water to yield a colorless solution.

When chlorine is added to this solution, the iodine is displaced since chlorine is above iodine in the electrochemical series.

When mixed with hexane, the color of the solution changes to purple.

5 0

3 years ago

ΔH for the reaction below is -826.0 kJ/mol. Calculate the heat change when a 69.03-g sample of iron is reacted.4Fe(s) + 3 O2(g)

Anuta_ua [19.1K]

Answer:

c. -1020.9 kJ

Explanation:

4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s) ΔH = -826.0 kJ/mol.

atomic weight of iron = 56

69.03 g = 69.03 / 56

= 1.23268 moles

Heat released by 1.23268 moles

= 1.23268 x 826.0

= -1020.9 kJ .

4 0

3 years ago

Which electromagnetic wave has the longest wavelength?

Vsevolod [243]

Radio waves has the longest wavelength

7 0

3 years ago

Read 2 more answers

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 $NaCl$