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Oxana [17]
4 years ago
9

When 200g water goes from a temperature 10C to 40C, is heat absorbed or released?

Chemistry
1 answer:
gayaneshka [121]4 years ago
8 0

how would it be possible to even be absorbed?? the answer is released.

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Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de
nignag [31]

Answer:

\%m/m=14\%

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14

Which is also the by-mass fraction and in percent it turns out:

\%m/m=0.14*100\%\\\\\%m/m=14\%

Best regards!

6 0
3 years ago
Determine the mass of MgCl2 needed to create a 100. ml solution with a concentration of 3.00 M.
kiruha [24]

Explanation:

1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\  = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1}  \: g \\  = 28.5 \: g \: of \: magnesium \: chloride

5 0
3 years ago
When iridium-192 is used in cancer treatment, a small cylindrical piece of 192ir, 0.6 mm in diameter and 3.5 mm long, is surgica
Ivan

Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?

Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.

V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3

Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":

9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)

= 6.96 x 10^19 atoms of Ir-122 are present.

4 0
3 years ago
Explain how the valence electrons are related to the element's period.
AlekseyPX

Answer:

The number of valence electrons in an atom is reflected by its position in the periodic table of the elements (see the periodic table in the Figure below). Across each row, or period, of the periodic table, the number of valence electrons in groups 1–2 and 13–18 increases by one from one element to the next

5 0
3 years ago
How many moles in 3.69 x 10^30 molecules of carbon dioxide?
Sedaia [141]

0.612 \times 10^{7} \text { moles of } \mathrm{CO}_{2} are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}

<u>Explanation:</u>

It is known that each mole of an element is composed of avagadro's number of molecules. So if we need to determine, we need to divide the number of molecules with the avagadro's number.

So,

    1 mol of element =6.02 \times 10^{23} molecules of element

As here 3.69 \times 10^{30} molecules of carbon di oxide is given. So the moles in it will be

   No. of moles of carbon dioxide = \frac{3.69 \times 10^{30}}{6.02 \times 10^{23}}

    No. of moles = 0.612 \times 10^{7} moles of carbon dioxide.

Thus,

0.612 \times 10^{7} of carbon dioxide are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}.

4 0
3 years ago
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