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NNADVOKAT [17]
3 years ago
9

What are the chemical properties of iron meta

Chemistry
1 answer:
kari74 [83]3 years ago
3 0

Answer:

Physical properties

Atomic mass: 55.845

Density: 7.87 g/cm3

Group: Transition metals

Electrons: [Ar] 3d6 4s2

Melting point: 1811 K

Boiling point: 3134 K

Discovered: Ancient times

Explanation:

hope this helps! :D

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Two charged particles repel each other with a force of F.
mihalych1998 [28]

Answer:

it would increase to twice what it was

3 0
3 years ago
The solubility table helps predict the products of what kind of reactions?
Ganezh [65]

Answer:

B. double replacement

Explanation:

Rate, or give brainliest if you can :)

3 0
2 years ago
28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir
Damm [24]
<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

         = 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

                             = 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

   = (0.0050 × 0.25)÷ 0.30

   = 0.004167 L or

   = 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0
3 years ago
(WILL MARK BRAINLIEST)<br> PLZ HELP ASAP
ArbitrLikvidat [17]

Answer:

a) 2NaOH(aq) + CuSO4(aq) -------------> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --------------> CaCO3 + H2O (this is already balanced)

c) Pb(NO3)2 + H2SO4 --------> PbSO4 + 2HNO3.

d) 2KNO3 ------> 2KNO2 + O2

e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

3 0
3 years ago
Read 2 more answers
what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

5 0
1 year ago
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