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3 years ago

Change 23 degrees Celsius to F

2 answers:

8 0

Rewrite the formula C=5/9(F-32) substituting 23 for C: 23=5/9(F-32), then multiply both sides by the reciprocal of 5/9.
(9/5)*(23)=(9/5)*5/9(F-32)
41.4=F-32; add 32 to both sides.
41.4+32=F-32+32
73.4=F

Marat540 [252]3 years ago

7 0

It's 73.5 F cz c/5=f-32/9

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Explain at least one observation you can make about the political borders and the language borders shown on the maps.

bazaltina [42]

Answer:

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7 0

2 years ago

At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the

Elis [28]

Answer:

P₂ = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂ = 25.83 atm .L. K / 293 K . L

P₂ = 0.09 atm

7 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

A 3.06 gram sample of an unknown hydrocarbon with empirical formula CH2O was found to contain 0.0170 moles of the substance. Wha

Yanka [14]

Answer:

180 amu

C₆H₁₂O₆

Explanation:

Step 1: Determine the molecular mass of the compound

The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:

M = m/n = 3.06/0.0170 mol = 180 g/mol

Then, the molecular mass is 180 amu.

Step 2: Determine the molar mass of the empirical formula.

M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)

M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol

Step 3: Determine the molecular formula

First, we will determine "n" according to the following expression.

n = molar mass molecular formula / molar mass empirical formula

n = 180 g/mol / 30 g/mol = 6

The molecular formula is:

n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

5 0

2 years ago

After mixing the solutions in a separatory funnel, the stopper should be ________and the liquid should be ________and the layers

sdas [7]

Answer:

The answer is below

Explanation:

The separation technique is used for separating immiscible liquids.

When separating, the stopper has to be removed when draining the lower layer so as to prevent a vacuum. If vacuum is allowed, the draining rate will reduce and stop.

The liquid should be mixed by shaking the funnel and then opening the stopcock so as the vent out gases.

When near interface between the layers, you should set your eye level so that you do not drain up to the second layer.

After completely draining the first layer, the second layer should be collected in a new flask.

After mixing the solutions in a separatory funnel, the stopper should be removed and the liquid should be mixed thoroughly and the layers allowed to separate. When you get close to the interface between the layers, get eye level with the funnel and slow the draining until the first layer is collected. Switch to a new flask to collect the second layer.

4 0

2 years ago