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3 years ago

Dude, What ⬆the b****direct dnt think thru jobs

Chemistry

1 answer:

sammy [17]3 years ago

7 0

Answer:

Just make it sound right please it don't even add up what was you even thinking like come on now...lol

Explanation:

What does this mean I don't understand your question.

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a solution of prepared by dissolving 20.0 brams of mavni fluride (MgF2) in 80.0 of water (H2O). whag is the mole fraction of mag

Serjik [45]

Answer:

2000.1234 trebuie sa fie asa

3 0

4 years ago

Assume that 25 ml of 0.440M NaCl are added to 25 ml of 0.320M AgNO3.

Genrish500 [490]

The balanced chemical equation would be as follows:

<span>NaCl + AgNO3 -> NaNO3 + AgCl

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl -----> consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3

0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced

</span>

8 0

3 years ago

Can some please answer these two questions?

alexandr402 [8]

Answer:

1. Mass of potassium (K) = 203.32 g

2. Number of mole of As = 7.53 moles

Explanation:

1. Determination of the mass of potassium (K)

Molar mass of K = 39.1 g/mol

Number of mole of K = 5.2 moles

Mass of K =.?

Mole = mass / Molar mass

5.2 = mass of K / 39.1

Cross multiply

Mass of K = 5.2 × 39.1

Mass of potassium (K) = 203.32 g

2. Determination of the number of mole of Arsenic (As)

Molar mass of As = 74.92 g/mol

Mass of As = 563.9 g

Number of mole of As =.?

Mole = mass /Molar mass

Number of mole of As = 563.9 / 74.92

Number of mole of As = 7.53 moles

8 0

3 years ago

You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1

jeka57 [31]

Answer:

$\boxed{\text{36 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O$

2. Calculate the moles of HCOOH

$\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}$

3. Calculate the moles of NaOH.

$\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}$

4. Calculate the volume of NaOH

$c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}$

3 0

3 years ago

Grtdggddgdggrsggrgrgd

n200080 [17]

Answer:

wut??

Explanation:

7 0

2 years ago