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3 years ago

Ultrasound is used to break up kidney stones. How do sound waves break up kidney stones?

Chemistry

1 answer:

dolphi86 [110]3 years ago

4 0

Answer:

eswl uses shock waves to break kidney stones into small pieces that can more easily travel through the urinary tract.

Explanation:

hope this helps

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What is the molarity of a solution that contains 1.67 moles in 1,750 mL?

solmaris [256]

Answer:

0.954M solution

Explanation:

molarity = moles solute / volume solution in liters

=> molarity = 1.67 moles / 1.750 L soln = 0.954M solution

8 0

2 years ago

0.254g lead(ii)ethanoate, on adding excess K2CrO4 solution, gave 0.130g of lead(ii)chromate precipitate. what is the percentage

Alborosie

Answer:

32.8%

Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄

There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:

4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb

Finally we calculate the percentage composition of Pb:

0.083 g Pb / 0.254 g salt * 100% = 32.8%

3 0

3 years ago

Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperat

KonstantinChe [14]

Answer:

$\boxed{\text{139 $\, ^{\circ}$C}}$

Explanation:

The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"

To answer this question, we can use the Clausius-Clapeyron equation:

$\ln \left (\dfrac{p_{2}}{p_{1}} \right) = \dfrac{\Delta_{\text{vap}}H}{R} \left(\dfrac{1 }{ T_{1} } - \dfrac{1}{T_{2}} \right)$

Data:

p₁ = 1 atm; T₁ = 373.15C

p₂ = 3.4atm; T₂ = ?

R = 8.314 J·K⁻¹mol⁻¹

$\Delta_{\text{vap}}H = \text{39.67 kJ//mol}$

(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)

Calculation:

$\begin{array}{rcl}\ln \left (\dfrac{p_{2}}{p_{1}} \right)& = & \dfrac{\Delta_{\text{vap}}H}{R} \left( \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right)\\\\\ln \left (\dfrac{3.4}{1} \right)& = & \dfrac{39670}{8.314} \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\\ln3.4 & = & 4771 \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\1.224 & = & 12.78 - \dfrac{4771}{T_{2}}\\\\\dfrac{4771}{T_{2}} & = & 11.56\\\\\end{array}$

$T_{2} & = & \dfrac{4771}{11.56} = \text{412.6 K} = \textbf{139 $\, ^{\circ}$C}\\\\\text{The maximum temperature is } \boxed{\textbf{139 $\, ^{\circ}$C}}$