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3 years ago

What oxidation number would you expect an iodide ion to have?

Chemistry

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

Explanation:

An iodide ion is the ion I−. Compounds with iodine in formal oxidation state −1 are called iodides.

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Water, as a polar molecule, has many interesting properties. Which of the following properties is most closely associated with p

beks73 [17]

all of the above.

The properties of water are:

1) polar molecules

2) regulates temperature

3) universal solvent

4) high surface tension

5) can exist in all three stages of matter

4 0

3 years ago

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The outer skin makes the surfboard more expensive. Suggest two reasons why an outer skin is added to the polystyerene core.

Vlada [557]

Answer:

It helps it not to break or crack. It is like an outer shell to protect it.

Explanation:

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3 years ago

HELP PLSSSSSSS IM SO DUMMMMMMMMMMMMMMMMMMMMM AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

vekshin1

Answer:

I'm sorry but I'm not doing the whole test

Explanation:

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3 years ago

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What is the reaction for CL^2 + 2 KBr —> 2 KCL+Br^2 of 11 grams of potassium bromide?

mylen [45]

Answer:

All the amounts of reactants and products are:

KBr

11.0g (given)

0.0924 mol

Cl₂

0.0462 mol

3.28g

KCl

0.0924 mol

6.89 g

Br₂

0.0462 mol

7.39 g

Explanation:

1. Balanced chemical equation (given)

$Cl_2+2KBr\rightarrow 2KCl+Br_2$

2. Mole ratios

$\dfrac{1molCl_2}{2molKBr}$

$\dfrac{2molKCl}{2molKBr}$

$\dfrac{1molBr_2}{2molKBr}$

3. Molar masses

Molar mass Cl₂: 70.906g/mol

Molar mass KBr: 119.002 g/mol

Molar mass KCl: 74.5513 g/mol

Molar mass KBr: 159.808 g/mol

4. Convert 11 grams of potassium bromide to moles:

#moles = mass in grams / molar mass

#mol KBr = 11g / 119.002g/mol = 0.092435mol KBr

5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂

a) Cl₂

$\dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2$

$0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2$

b) KCl

$\dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl$

$0.092435molKCl\times 74.5513g/molKCl=6.89gKCl$

c) Br₂

$\dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2$

$0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2$

The final calculations are rounded to 3 sginificant figures.

3 0

3 years ago

A gas occupying a volume of 656.0 mL at a pressure of 0.884 atm is allowed to expand at constant temperature until its pressure

7nadin3 [17]

Answer:

1.14 × 10³ mL

Explanation:

Step 1: Given data

Initial volume of the gas (V₁): 656.0 mL

Initial pressure of the gas (P₁): 0.884 atm

Final volume of the gas (V₂): ?

Final pressure of the gas (P₂): 0.510 atm

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL

5 0

3 years ago