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Bumek [7]
1 year ago
14

From the following list of elements, those that will always form ionic compounds in a 1:2 ratio with zinc.

Chemistry
1 answer:
harina [27]1 year ago
3 0

Answer:

A. iodine

C. fluorine

F. bromine

Explanation:

Ionic bonds occur mostly between metals and non-metals. Usually, a wide electronegativity difference is preferred between the two atoms. This makes one atom more desirous to gain electron and other more willing to donate electrons.

To have Zn forming  a compound in the ratio of 1 to 2, the combining power must be similar to this.

The dominant oxidation state of Zn is the is the +2 state.

The other combining atoms must have the ability to recieve the two electrons.

The halogens fit perfectly into this picture. They need just an electron to attain nobility. They are also highly electronegative. If two halogens combines with the Zn, then the ionic bond will result.

The halogens are fluorine,chlorine, bromine, iodine and astatine.

They will form these compounds:

                       ZnF₂, ZnBr₂ and ZnI₂

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Calculate the mass defect of the boron nucleus 11 5b. The mass of neutral 11 5b is equal to 11.009305 atomic mass units.
Vlad1618 [11]

Answer:

0.07906687 amu

Explanation:

For Boron ₅B¹¹, the number of protons is 5 and the mass is 11. The mass is the number of protons plus the number of neutrons, so:

neutrons = 11 - 5 = 6

The mass of an atom is concentrated in the nucleus, so it is the mass of the protons + the mass of the neutrons. The mass of 1 proton is 1.00727647 amu/proton, and the mass of 1 neutron: 1.00866492 amu/neutron, so for the element given the theoretical mass (mt) is:

mt = 5* 1.00727647 amu/proton + 6*1.00866492 amu/neutron

mt = 11.08837187 amu

The mass defect (md) is the theorical mass less the real mass:

md = 11.08837187 - 11.009305

md = 0.07906687 amu

7 0
2 years ago
Read 2 more answers
What is the number of moles in 15.0 g AsH3?
Grace [21]

Answer:

0.192 mol.

Explanation:

  • To calculate the no. of moles of a substance (n), we use the relation:

<em>n = mass / molar mass.</em>

mass of AsH₃ = 15.0 g.

molar mass of AsH₃ = 77.95 g/mol.

∴ The number of moles in 15.0 g AsH₃ = mass / molar mass = (15.0 g) / (77.95 g/mol) = 0.192 mol.

7 0
1 year ago
3. A mass of 0.15 ounces is equal to how many grams?
garik1379 [7]

Answer:

Option C = 4.25 g

Explanation:

Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.

Given data:

Mass = 0.15 ounce

Mass in gram = ?

Solution:

One ounce is equal to 28.35 g, so

0.15 × 28.35 = 4.25 g

5 0
2 years ago
The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d
OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

5 0
2 years ago
A food product is being frozen in a system capable of removing 6000 kJ of thermal energy. The product has a specifi c heat of 4
lapo4ka [179]

Answer : The exit temperature of the product is, -35.2^oC

Explanation :

Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen

Q=m\times c_1\times (T_2-T_1)+m\times L_f+m\times c_2\times (T_4-T_3)

where,

Q =  Total heat = 6000 kJ

m = mass of product = 15 kg

c_1 = specific heat of liquid = 4kJ/kg^oC

L_f = latent heat of fusion = 275kJ/kg

c_2 = specific heat of frozen = 2.5kJ/kg^oC

T_1 = initial temperature of liquid = 2^oC

T_2 = final temperature of liquid = 10^oC

T_3 = initial temperature of frozen = ?

T_4 = final temperature of frozen = 2^oC

Now put all the given value in the above expression, we get:

6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC]

T_3=-35.2^oC

Thus, the exit temperature of the product is, -35.2^oC

7 0
1 year ago
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