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3 years ago

How many grams of Cl are in 535 g of CaCl₂?

1 answer:

7 0

The answer is 341.7 g.

(1) Calculate the molar mass (M) of CaCl2 which is the sum of atomic masses (A) of elements:
M(CaCl2) = A(Ca) + 2A(Cl)
A(Ca) = 40.1 g/mol
A(Cl) = 35.45 g/mol

M(CaCl2) = 40.1 + 2 * 35.45 = 40.1 + 70.9 = 111 g/mol

(2) Calculate in how many moles are 535 g:
M(CaCl2) = 111 g/mol
111g : 1mol = 535 g : xmol
x = 535 g * 1mol : 111g = 4.82 mol

(3) Calculate how many grams of Cl are in 4.82 mol:
A(Cl) = 35.45 g/mol
2A(Cl) = 2 * 35.45 = 70.9 g/mol
70.9 g : 1 mol = x : 4.82 mol
x = 70.9 g * 4.82 mol : 1 mol = 341.7 g

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Convert the following: 100 hg(hectograms) to g(grams)

solniwko [45]

Answer:

Explanation:

10,000g

just multiply the amount of hectograms by 100 to get grams

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4 years ago

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N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?

BartSMP [9]

Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.

Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.

Oxidation number of H on product side is 0 as it is present in its elemental for, $H_2$ ,

So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.

4 0

3 years ago

If the density of pure water is 0.9922 g/mL at 40 ºC, calculate its theoretical molarity at that temperature. Report to 4 sig fi

OleMash [197]

Answer is: theoretical molarity of water is 55.1222 mol/L.
d(H₂O) = 0.9922 g/mL.
M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.
M(H₂O) = 2 + 16 · g/mol = 18 g/mol.
c(H₂O) = d(H₂O) ÷ M(H₂O).
c(H₂O) = 0.9922 g/mL ÷ 18 g/mol.
c(H₂O) = 0.0551 mol/mL.
c(H₂O) = 0.0551 mol/mL · 1000 mL/L = 55.1222 mol/L.

3 0

4 years ago

How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?

Allisa [31]

We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

0.266 M x V1 = 0.075 M x 150 mL

V1 = 42.29 mL

Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.

3 0

3 years ago

Read 2 more answers

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago