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mart [117]
3 years ago
11

Match each lab incident to the item of PPE that will protect you from it. Each item of PPE will only be used once.

Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

fefeff

Explanation:

wdwdwfefe

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If 100mL was diluted of a 2.0 M solution of sodium chloride by adding an additional 100 mL of water, what would the new molarity
andrew-mc [135]
I need help with my questions sorry
7 0
2 years ago
A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

4 0
3 years ago
When the gases dihydrogen monosulfide and oxygen react, they form the gases sulfur dioxide and water vapor. Write the balanced c
Anna35 [415]
2(H2S) + 3(O2) -> 2(SO2) + 2(H2O)
7 0
2 years ago
"Calculate the number of millimoles in 500 mg of each of the following amino acids: alanine (MW = 89), leucine (131), tryptophan
Cloud [144]

Answer:

- Alanine =  5.61 mmoles

- Leucine = 3.81 mmoles

- Tryptophan = 2.45 mmoles

- Cysteine = 4.13 mmoles

- Glutamic acid = 3.40 mmoles

Explanation:

Mass / Molar mass = Moles

Milimoles = Mol . 1000

500 mg / 1000 = 0.5 g

- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles  . 1000 = 5.61mmoles

- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles  . 1000 = 3.81 mmoles

- Tryptophan = 0.5 g / 204 g/m →  2.45x10⁻³ moles . 1000 = 2.45 mmoles

- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles

- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 =  3.4 mmoles

5 0
3 years ago
What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?
hjlf

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v =  1560 g / 41.2 g/ml

v = 37.9 ml

7 0
3 years ago
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