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3 years ago

Guys please I need an answer for this asap

1 answer:

6 0

I need more information, there are meant to be coefficients in front of one of those ..

but you’d put .2 over the coefficients, and if there is not one, you put an X. then you will cross multiply, and divide if you have 2x or something greater than X.

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i’m in desperate need of help haha. btw it’s 200 ml of a clear liquid and 50 ml of a different clear liquid. please help

Lelu [443]

C. a chemical reaction occurred when the two liquids were mixed

6 0

4 years ago

Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

Aliun [14]

Answer:

1.60.

Explanation:

The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.

The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

∴ pH = - log[H⁺] = - log(0.025) = 1.602 ≅ 1.60.

5 0

4 years ago

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

3 0

3 years ago

the combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how many moles of water formed

enyata [817]

$2C_4H_{10}+13O_2$ ⇒ $8CO_2 + 10H_2O$

$n= \frac{m}{M} = \frac{m}{2M(H)+M(O)}= \frac{2,46}{2*1,0+16,0} = 0,14mol$

So 0,14mol are formed.

6 0

4 years ago

The weak base ionization constant (Kb) for HCO3 is equal to:

hammer [34]

Weak bases are alkaline solution that does not get completely dissociated. The dissociation constant will be the ratio of the concentration of the products to the reactants.

<h3>What is a weak base ionization constant?</h3>

The weak base ionization constant is the equilibrium constant that is given as the division of the products of the ionization to the concentration of the reactants.

The reaction for HCO₃ is given as:

HCO₃⁻ + OH⁻ ⇄ CO₃²⁻ + H₂O

The value of Ka for the given reaction will be:

Ka = [CO₃²⁻][H₂O] ÷ [HCO₃⁻ ][OH⁻]

Therefore, the Ka of the weak carbonic acid is given as the ratio of the concentration of the products to the reactants.

Learn more about the weak ionization constant here:

brainly.com/question/27200344

#SPJ1

8 0

2 years ago