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3 years ago

Which on goes where please help!!!

Chemistry

1 answer:

EastWind [94]3 years ago

6 0

Pretty easy.. The ice is a solid, the rain is a liquid the drink is a liquid, the water drain is a liquid, the wood is a solid,

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A chemist must prepare 800.0mL of sodium hydroxide solution with a pH of 12.10 at 25°C. She will do this in three steps: Fill a

AnnZ [28]

Answer:

0.42 g

Explanation:

We have:

pH = 12.10 (25 °C)

V = 800.0 mL = 0.800 L

To find the mass of sodium hydroxide (NaOH) we can use the pH:

$14 = pH + pOH$

$pOH = 14 - pH = 14 - 12.10 = 1.90$

$pOH = -log ([OH^{-}])$

$[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M$

Now, we can find the number of moles (η) of OH:

$\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles$

Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.

Finally, with the number of moles we can find the mass of NaOH:

$m = \eta * M$

Where M is the molar mass of NaOH = 39.9 g/mol

$m = 1.04 \cdot 10^{-2} moles * 39.9 g/mol = 0.42 g$

Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.

I hope it helps you!

3 0

3 years ago

Methane (CH4) reacts with excess oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). What is the percent yield of c

kotegsom [21]

I'm pretty sure that it's either 83.2% or 91.4%

6 0

3 years ago

For the reaction C (s) + H2O (g) --> CO (g) + H2 (g) ΔH = 131.3 kJ/mol and ΔS = 133.6 J/K-mol at 298K. At temperatures greate

TiliK225 [7]

Answer:

The reaction is spontaneous when T> 0.98 Kelvin OR T> -272.17°C

Explanation:

Step 1: Data given

ΔH = 131.3 kJ/mol = 131300 J/mol

ΔS = 133.6 J/K*mol

T = 298K

Step 2: The balanced equation

C (s) + H2O (g) --> CO (g) + H2 (g)

Step 3: ΔG

For a reaction to be spontaneous, ΔG should be <0

When ΔG > 0 the reaction is spontaneous in the reverse direction.

ΔG = ΔH - TΔS

Since ΔG<0

ΔH - TΔS <0

Step 4: Calculate T where the reaction is spontaneous

ΔH - TΔS <0

131300 J/mol - T*133.6 J/K*mol <0

- T*133.6 J/K*mol < -131300 J/mol

-T <-131300 /133.6

-T< -982.8 Kelvin

T> 982.8 Kelvin OR T> 709.6°C

The reaction is spontaneous when T> 982.8 Kelvin OR T> 709.6°C

At 298 K this reaction C (s) + H2O (g) --> CO (g) + H2 (g) is not spontaneous

6 0

3 years ago

Read 2 more answers

Which of these can be classified as a pure substance

Eduardwww [97]

Salt can be classified as a pure substance as it is only composed of sodium chloride
while contain many other substances

8 0

3 years ago

Read 2 more answers

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

2 years ago