All categories

2 years ago

ANYONE PLEASE HELP ME WITH MY CHEMISTRY HOMEWORK I REALLY NEED THE ANSWER RIGHT NOW I HOPE Y’ALL CAN HELP ME:(

Chemistry

1 answer:

stealth61 [152]2 years ago

6 0

2. They all have 1 valence electron, they form ions with a +1 charge.

3. cation

4. Lithium, Li

5. Francium, Fr

6. Cesium has a melting point of 28.5 C so I’m assuming that’s the correct answer, Francium is radioactive

7. Sodium, Rubidium, Lithium, and Cesium are all soft metals that can be cut with a knife.

8. Oxygen

9. Hydrogen gas

10. Bases

You might be interested in

If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?

photoshop1234 [79]

The combustion reaction is as expressed,

CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.

moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.

7 0

3 years ago

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(d) $1s^22s^22p^63s^23p^63d^4s^1$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^63d^4s^1$ → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: $(n-1)d^{1-10}ns^{0-2}$ where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^63d^4s^1$

The element having this electronic configuration belongs to the transition family.

4 0

3 years ago

How much energy is needed to change 44 grams of ice at -8° to steam at 107°?

Tpy6a [65]

Q=m(c∆t +heat of fusion + heat of evaporation)

m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g

Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J

5 0

3 years ago

\"Moving down group 2A (Alkaline Earth Metals), which element has the largest first ionization energy?\" Is the answer to this:

lesya692 [45]

Ionization energy is the energy required to remove the outermost electron from one mole of gaseous atom to produce 1 mole of gaseous in to produce a charge of 1. The greater the ionization energy, the greater is the chance f the electron to be removed from the nucleus. In this casse, Radium has the largest ionization energy.

6 0

3 years ago

"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi

ycow [4]

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

7 0

4 years ago