Name the following compounds CaSe don’t use common names such as water or salt

Consider the word equation. Calcium hydroxide hydrochloric acid mc012-1. Jpg calcium chloride water Which is the corresponding f

dimulka [17.4K]

The corresponding formula equation is:

Upper Ca (upper OH) subscript 2 (s) plus 2 upper H upper Cl (l) right arrow upper Ca upper Cl subscript 2 (aq) plus 2 upper H subscript 2 upper O (l)

To obtain the word equation, we shall write out the balanced molecular equation. This is given below:

Calcium hydroxide => Ca(OH)₂

Hydrochloric => HCl

Calcium chloride => CaCl₂

Water => H₂O

Calcium hydroxide + Hydrochloric —> Calcium chloride + water

Ca(OH)₂ + HCl —> CaCl₂ + H₂O

There are 2 atoms of Cl on the right side and 1 atom on the left. It can be balance by writing 2 before HCl as shown below:

Ca(OH)₂ + 2HCl —> CaCl₂ + H₂O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 2 before H₂O as shown below:

Ca(OH)₂ + 2HCl —> CaCl₂ + 2H₂O

Now, the equation is balanced.

Thus, the corresponding word equation is:

Upper Ca (upper OH) subscript 2 (s) plus 2 upper H upper Cl (l) right arrow upper Ca upper Cl subscript 2 (aq) plus 2 upper H subscript 2 upper O (l)

Learn more about balancing equation: brainly.com/question/15538713

8 0

2 years ago

Read 2 more answers

g 10.00 mL of phosphoric acid (H3PO4) are titrated with 0.1000 M sodium hydroxide. 15.00 mL of the sodium hydroxide solution are

aliya0001 [1]

Answer: The molarity of phosphoric acid is 0.05 M

Explanation:

A neutralization reaction is defined as the reaction in which an acid reacts with a base to form a salt and water molecule.

At equivalence, moles of acid becomes equal to the moles of base. The equation for the neutralization follows:

$n_1M_1V_1=n_2M_2V_2$ ........(1)

where,

$n_1, M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid that is $H_3PO_4$

$n_2, M_2\text{ and }V_2$ are the n-factor, molarity and volume of the base that is NaOH

Given values:

$n_1=3\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.1000M\\V_2=15.00mL$

Plugging values in equation 1:

$3\times M_1\times 10.00=1\times 0.1000\times 15.00\\\\M_1=\frac{1\times 0.1000\times 15.00}{3\times 10.00}\\\\M_1=0.05M$

Hence, the molarity of phosphoric acid is 0.05 M

6 0

3 years ago

What are the two half-reactions that occur in this electroplating process. Be sure to include the states of the elements or ions

Serhud [2]

This is rather a case of purification of impure copper or extraction of Cu from its alloy. You need to place the impure copper rod on the positive electrode (Anode) usualyy made of carbon rod, whereby oxidation reaction takes place: Cu (s) -------> Cu2+ (aq) + 2e- That is to say the impure cooper dissolves into solution. The copper (II) ions move to the negative electrode (cathode) usually made of pure copper rod. At the cathode, the Cu2+ ions are reduced : Cu2 (aq)+ + 2e- -----> Cu(s). That is to say the copper (II) ions are deposited as solid copper atoms onto the cu-rod electrode. In this way impure copper is deposited as pure copper onto the copper cathode

4 0

3 years ago

What s thisssss oneeeeeee

ZanzabumX [31]

C
Explanation- equals 459

3 0

3 years ago

i need to kono why this app is so ....i get a add then i hav ea nohter add and there way to long. anyways 3 more weekds till sum

erastova [34]

Answer:

same I hate all the frickin ads ;-;

you gave a good day too mate