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3 years ago

What is the uncertian digit in 0.054 g

Chemistry

1 answer:

UkoKoshka [18]3 years ago

8 0

Answer:

5.4g

Explanation:

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WILL MARK BRAINLIEST FOR CORRECT ANSWER

jenyasd209 [6]

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4 years ago

Read 2 more answers

Write and balance the equation for the dissolving of aluminium chloride into an aqueous solution with the dissociation of

umka2103 [35]

The balanced equation for the dissolving of aluminum chloride into an aqueous solution with the dissociation of the ions is; AlCl3(s) → Al^3+ (aq) + 3 Cl^-(aq).

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The equation that shows the balanced equation for the dissolving of aluminum chloride into an aqueous solution with the dissociation of the ions is; AlCl3(s) → Al^3+ (aq) + 3 Cl^-(aq).

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6 0

2 years ago

Helpppp helpppp plsssss

Sedbober [7]

$\Large \text{1)}$

$\textbf{Use the formula and substitute the values}$

$\displaystyle 0.3\ mol/L = \frac{n \ mol}{10 \times 10^{-3}\ L}$

$\textbf{Solve for the number of moles}$

$n = 0.003\ mol$

$\textbf{Calculate final volume }$

$40+10=50\ mL$

$\textbf{Number of moles remains same. Solve for concentration}$

$\displaystyle C=\frac{0.003\ mol}{50 \times 10^{-3}\ L} =0.06\ mol/L$

$\Large \text{2)}$

$\displaystyle DF=\frac{V_f}{V_i}$

$\displaystyle DF=\frac{50}{10}=5$

3 0

3 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

How many moles of 18L of NH3 at 30 celcius and 912 mmHg? how many grams?

babymother [125]

Answer is: mass of ammonia is 14.76 grams and amount is 0.87 moles.<span>
V(NH</span>₃<span>) = 18 L.
T = 30°C = 303.15 K.
p = 912 mmHg ÷ 760 mmHg/atm= 1.2 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n</span> = p·V / R·T.
n(NH₃) = 1.2 atm ·18 L / 0.08206 L·atm/mol·K · 303.15 K.
n(NH₃) = 0.87 mol.
m(NH₃) = n(NH₃) · M(NH₃).
m(NH₃) = 0.87 mol · 17 g/mol.
m(NH₃) = 14.76 g.<span>

</span>

6 0

4 years ago