Answer:
96 g
Explanation:
Tomando en cuenta que estamos hablando de una solución de Nitrato de potasio cuya concentración es 12% m/V, debemos entonces usar la expresión de %m/V:
%m/V = msto / Vsol * 100 (1)
De esta expresión, debemos despejar la masa de soluto (msto) que es precisamente el valor que estamos buscando. Al hacer el despeje tenemos:
msto = %m/V * Vsol / 100 (2)
Ahora solo nos queda sustituir los valores dados y resolver:
msto = 12 * 800 / 100
<h2>
msto = 96 g de KNO₃</h2>
Espero te sirva.
Answer:
Hi, a cheerful teen willing to help,
wishing you a splendiferous day ahead...
Explanation:
1ai. The most dense is Tungsten.
1aii. The least dense is Sodium.
1aiii. The strongest is Tungsten.
1b. All metals conduct electricity.
They are all magnetic except magnetic.
and all solid at room temperature except mercury.
1c. Mercury is a liquid at room temperature.
1d. Tungsten is used as filament for light bulbs due to its high melting point and it doesn't oxidize, this the filament won't melt.
1e. The lump of gold will sink because it is more dense than mercury.
Answer:
B
Explanation:
It is a gas and therefore according to particle model of matter it has large spaces between them and does not have a definite shape
Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
