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3 years ago

10

Calculate the molarity of 1.60L of a solution containing 1.55g of dissolved KBr

1 answer:

Olegator [25]3 years ago

7 0

Answer: the molarity is .00814

Explanation:the formula for molarity is moles over liters but we have grams so we do given over one so 1.55 over 1 multiplied by 1/119.0023 which is tht two atomc masses addes together then you get the moles and divde them by the liters to get the molarity hope this helps god bless

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58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2,

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<em><u>Answer and Explanation:</u></em>

$Greetings!$

$Let's~answer~your~question!$

$Partial ~pressure ~of ~gas ~can ~be ~directly~ calculated ~by ~multiplying ~the~ percentage\\ of~ pressure~ of~ gases~ to ~the ~total ~pressure.$

$\boxed{Pgas~ = ~P~total~ * \% ~P ~of ~gas}$

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3 years ago

In a strong acid–strong base titration (both monoprotic), if 25.0 milliliters of the base is required to completely neutralize 2

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Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were intr

dalvyx [7]

Explanation:

The given data is as follows.

$V_{1}$ = 250 mL, $V_{2}$ = 750 mL

$T_{1}$ = $35^{o}C$ = 35 + 273 K = 308 K

$T_{2}$ = 35 + 273 K = 308 K

$P_{1}$ = 0.55 atm, $P_{2}$ = 1.5 atm

P = ? , V = 10.0 L

Since, temperature is constant.

So, $P_{1}V_{1} + P_{2}V_{2}$ = PV

Now, putting the given values into the above formula as follows.

$P_{1}V_{1} + P_{2}V_{2}$ = PV

$0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm$ = $P \times 10.0 L$

P = 0.126 atm

As, 1 atm = 760 torr. So, $0.126 atm \times \frac{760 torr}{1 atm}$ = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.

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