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2 years ago

The bonds in the compound MgSO4 can be described as

1 answer:

7 0

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

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The standard reduction potentials for two half-cells involving iron are: given below. Fe2+ (aq) + 2e– ® Fe (s) Eο = –0.44 V Fe3+

matrenka [14]

Answer:

Explanation:

Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V

Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V

Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .

So reaction in the combined cell will be

2Fe⁺³ + Fe = 3Fe⁺²

cell potential = .77 - ( - .44 )

= 1.21 V .

6 0

2 years ago

Be sure to answer all parts. calculate δg o for the reaction between i2(s) and br−(aq). e o cell = −0.54 j/c enter your answer i

denpristay [2]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

The attached file have the solved problem.

3 0

2 years ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago

Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t

Gwar [14]

Answer

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

x = 5

7 0

2 years ago

6. Determine the amount in moles of the following:<br>a. 12.15 g Mg<br>b. 1.50 x 1023 atoms F

igomit [66]

a. 0.51 moles of Mg

b. 0.25 moles of F

Explanation:

a. To find the number of moles knowing the mass we use the following formula:

number of moles = mass / molecular weight

number of moles of Mg = 12.15 / 24 = 0.51 moles

b. To find the number of moles knowing the number of atoms we use Avogadro's number to illustrate the following reasoning:

if in 1 mole of F there are 6.022 × 10²³ atoms of F

then in X moles of F there are 1.5 × 10²³ atoms of F

X = (1 × 1.5 × 10²³) / 6.022 × 10²³

X = 0.25 moles of F

Learn more about:

Avogadro's number

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3 0

1 year ago