HELP ME ASAP!!!!!!!!

Global climate change is

WARRIOR [948]

It’s d man..........

4 0

3 years ago

What is the molarity of a solution that contains 4.3 mol of acetic acid (C2H3O2) in 450. mL of solution?

lord [1]

Hey there !

Number of moles of solution: 4.3 moles

Volume in liters:

450.0 mL / 1000 => 0.45 L

Therefore:

Molarity = number of moles / volume of solution ( L)

Molarity = 4.3 / 0.45

=> 9.55 M

Hope that helps!

3 0

2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

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

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

2 years ago

Read 2 more answers

a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element

Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0

3 years ago

What is the % composition of Carbon in Chromium (iii) Carbonate

photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

$Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3$

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

$\begin{gathered} C\rightarrow3\times12=36 \\ \\ O\rightarrow9\times16=144 \\ \\ Cr\rightarrow2\times52=104 \end{gathered}$

The molar mass will be thus:

$M=36+104+144=284\text{ g/mol}$

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

$\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\ \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}$

The percent composition of Carbon is thus 12.7 %.

8 0

1 year ago