Help science!! *10 points*

What is the concentration of a 1:10 dilution of a 2.5 M solution of NaCl? How would you prepare exactly 100 ml of such a solutio

Marina CMI [18]

<u>Answer:</u> The concentration of the solution is 0.25 M

<u>Explanation:</u>

Let the volume of solution of 2.5 M NaCl be 10 mL

We are given:

Dilution ratio = 1 : 10

So, the solution prepared will have a volume of = $\frac{10}{1}\times 1000=100mL$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated NaCl solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted NaCl solution

We are given:

$M_1=2.5M\\V_1=10mL\\M_2=?M\\V_2=100mL$

Putting values in above equation, we get:

$2.5\times 10=M_2\times 100\\\\M_2=\frac{2.5\times 10}{100}=0.25M$

Hence, the concentration of the solution is 0.25 M

6 0

3 years ago

Twenty irregular pieces of an unknown metal have a collective mass of 28.225 g. When carefully placed in a graduated cylinder th

Lelechka [254]

Explanation:

Use the density formula to determine the volume of the piece of metal.

density

=

mass

volume

Rearrange the equation to isolate volume.

volume

=

mass

density

volume

=

147

g

7.00

g

mL

=

21.0 mL

The final volume in the cylinder after adding the piece of metal is

20.0 mL

+

21.0 mL

=

41.0 mL

8 0

3 years ago

If a measurement is said to be precise this means that it is what

Gala2k [10]

On point? Do you have any options?

6 0

3 years ago

Read 2 more answers

What is the volume of 60 g of ether if the density of ether is 70 g/mL

NNADVOKAT [17]

V = 60.0 g/ 0.70 g/mL = 85.7 mL Hope this helps! ;D

4 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

3 years ago