Balance the equation Ca(s) + H3PO4(aq) ---->Ca3(PO4)2(s) + H2(g)

What is the pH of a 0.100 M HI solution? Is that neutral, acidic, or basic? acidic What is the pOH of a 0.100 M HI solution? Is

inessss [21]

Answer: The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

Explanation:

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

The pH range of acidic solution is 0 to 6.9
The pH range of basic solution is 7.1 to 14
The pH of neutral solution is 7.

We are given:

Concentration of HI = 0.100 M

1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=0.100M$

Putting values in above equation, we get:

$pH=-\log(0.100)\\\\pH=1$

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14

$pOH=14-1=13$

Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

4 0

3 years ago

Is using gasoline to drive a chemical or physical

Flauer [41]

Answer:

Explanation:

Driving a car (burning gas is a chemical change) and almost all the plastics we use are made by chemical reactions of different components.

3 0

3 years ago

If m = 45g and V = 15ml, what is D in g/ml?

ch4aika [34]

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

6 0

3 years ago

What are two properties of most nonmetals

vekshin1

Answer: Summary of Common Properties

High ionization energies.
High electronegativities.
Poor thermal conductors.
Poor electrical conductors.
Brittle solids—not malleable or ductile.
Little or no metallic luster.
Gain electrons easily.
Dull, not metallic-shiny, although they may be colorful.

Explanation:

May I please have brainiest?

3 0

3 years ago

Read 2 more answers

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

3 years ago