Answer: option <span>A) increases from bottom to top within the group.
Explanation:
</span>It is a known trend that the metallic character of the elements increase from let to right and from top to bottom.
The greater the metallic character the greater the reactivity of the metal.
So, the elements of the columns 1 and 2 are the most reactive metals and among them the elements at the bottom are yet more reactive.
<span>The higher reactivity of the metals that are lower in the periodic table is attributed to the greater total number of electrons.
The greater the total number of electrons the more reactive the metals
as their outermost electrons (the valence electrons which are those that react) are located further from the nucleus and therefore they are held less
strongly, which makes them react more easily.</span>
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³
By definition,
Density = Mass/Volume
= (51.1 g)/(6.63 cm³)
= 7.7074 g/cm³
In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
= 7707.4 kg/m³
Answer: 7.707 g/cm³ or 7707.4 kg/m³
Answer:
Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form.
Explanation:
Chemical change involves formation of new substances. Therefore, the correct answer is "Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form."
Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
Best regards!
Answer:
CaCl2
Explanation:
As temperature increases pressure also increases so when the temperature is at 25 degrees C CaCl2 is already soluable in 100g of h20
