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2 years ago

An observation that is written as a number or amount?

Chemistry

1 answer:

beks73 [17]2 years ago

3 0

Answer: A quantitative observation

Explanation: Quantitative observations deal with numbers or amounts. As opposed to <em>qualitative</em> observations, which are observations made with the 5 senses.

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A sample of a substance has a mass of 4.2 grams and a volume of 6 milliliters

allsm [11]

Answer:

Density of the substance is 0.7

Explanation:

4.2/6 = 0.7

6 0

3 years ago

Read 2 more answers

Limiting Reactant Worksheet

faltersainse [42]

Answer:

a) 1.61 mol

b) Al is limiting reactant

c) HBr is in excess

Explanation:

Given data:

Moles of Al = 3.22 mol

Moles of HBr = 4.96 mol

Moles of H₂ formed = ?

What is limiting reactant =

What is excess reactant = ?

Solution:

Chemical equation:

2Al + 2HBr → 2AlBr + H₂

Now we will compare the moles:

Al : H₂

2 : 1

3.22 : 1/2×3.22 = 1.61 mol

HBr : H₂

2 : 1

4.96 : 1/2×4.96 = 2.48 mol

The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.

Moles of H₂ :

Number of moles of H₂ = 1.61 mol

6 0

2 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few

OlgaM077 [116]

Answer:

Incomplete precipitation of barium sulfate

Explanation:

The student has precipitated and digested the barium sulfate on his/her side. But on the addition of $BaCl_2$ in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When $BaCl_2$ is added, there are still sulfate ions present in the solution with combines with $BaCl_2$ and forms $BaSO_4$ and the formation of this precipitate makes the solution cloudy.

7 0

3 years ago

Can someone helps me

Lorico [155]

Answer:

Letter C is the correct one

6 0

2 years ago