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natta225 [31]
3 years ago
5

When working with a volumetric pipet, one adjusts the liquid level?

Chemistry
1 answer:
Lilit [14]3 years ago
7 0
One adjust the pipet by releasing the air to lower the pressure and equilizing the liquid level
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Which elements are<br> considered "Noble Metals"?
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ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).

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Which of the following elements would form the smallest ionic radius and why?
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2 years ago
The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

7 0
2 years ago
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

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3 years ago
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