All categories

2 years ago

How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH

1 answer:

4 0

To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
Liters= 0.0010 Liters ---------------->>>>>>>>>> 1.0 mL= 0.0010 Liters

moles= 2.0 M x 0.0010 Liters= 0.0020 moles

You might be interested in

Can someone answer that first this pls im lost

Alchen [17]

-Just look up “H2O lewis structure
-1.5
-Don’t know the VSEPR
-Polar Covalent
-Again, don’t know VSEPR
-Just look up H2O molecule

4 0

2 years ago

Arrange these elements based on their atomic radii. ga, f, s, as

mamaluj [8]

please have look at Periodic table , you will solve it yourself !

5 0

2 years ago

What volume of 1.10 M SrCl2 is needed to prepare 525 mL of 5.00 mM SrCl2?

grigory [225]

Answer:- 2.39 mL are required.

Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:

$M_1V_1=M_2V_2$

Where, $M_1$ and $M_2$ are molarities of concentrated and diluted solutions and $V_1$ and $V_2$ are their respective volumes.

$M_1$ = 1.10M

$M_2$ = 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)

$V_1$ = ?

$V_2$ = 525 mL

Let's plug in the given values in the formula:

$1.10M(V_1)=0.005M(525mL)$

$V_1=(\frac{0.005M*525mL}{1.10M})$

$V_1=2.39mL$

So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.

7 0

2 years ago

A change in matter that produces one or more new substances is a(n)

lukranit [14]

You could use another word for change can be variable witch means change and if you times the one two more times then you would get four because two time two would be four and times the one would be four.

6 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$