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3 years ago

Select the correct location on the image.

Chemistry

1 answer:

Mice21 [21]3 years ago

7 0

Answer:

The third one from the left–the graduated cylinder.

Explanation:

The laboratory apparatus that gives an "accurate" or "precise" measurement of a liquid's volume is the graduated cylinder. All you have to do is to pour the liquid into the cylinder and read its measurement using the calibrated scale.

The graduated cylinder comes in different sizes, which means the scale divisions will depend on its size. When reading the measurement, it is important to take note to read at the bottom of the meniscus because it gives the most accurate volume.

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Choose all the answers that apply.

Lady_Fox [76]

It is softer than topaz and and it is softer than diamond (diamond has a Mohs hardness of 10, which is the highest value of the scale)

7 0

3 years ago

Read 2 more answers

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)

Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0

3 years ago

A solution at 25 degrees Celslus 1.0*10^ -5 MH 3 O^ + . What the concentration of OH^ - in this solution?

dexar [7]

Answer:

Explanation:

-log(1.0x10^-5) = pH

pH + pOH = 14 (rearrange it)

OH- = 10^-pOH = 1.0 x 10^-9

- Hope that helped! Let me know if you need further explantion.

7 0

3 years ago

When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the freezing point of the resulting solution is 3

Lelu [443]

Answer: 120g/mol

Explanation:

The first step we are to take is to calculate the freezing point depression of the solution.

ΔT(f) = freezing point of pure solvent - freezing point of solution

ΔT(f) = 5.48 - 3.77

ΔT(f) = 1.71°C

Next we are to calculate the molal concentration of the solution using freezing point depression

ΔT(f) = K(f) * m

m = ΔT(f)/K(f)

m = 1.71/5.12

m = 0.333 molal

Now, we calculate the molecular weight of the unknown...

m = 0.333 mol = 0.333 mol X per kg of benzene

moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene

moles of X = 0.1665

molecular weight of X = 20g of X/0.1665

molecular weight of X = 120/mol

5 0

4 years ago