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strojnjashka [21]
2 years ago
14

The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of

99.8 kilopascals, what is the gauge pressure inside the container? A. 1.5 kPa B. 24.1 kPa C. 25.6 kPa D. 112.6 kPa E. 225.2 kPa
Chemistry
1 answer:
AlexFokin [52]2 years ago
4 0

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

Where,

PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

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A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft
luda_lava [24]

Answer:

42.5 g

Explanation:

Calculate the mass of the soft drink given the density and volume:

355 mL × 1.04 g/mL = 369.2 g

Now calculate the mass of sucrose given the percentage:

0.115 × 369.2 g = 42.46 g

Rounded to 3 significant figures, the mass is 42.5 g.

5 0
2 years ago
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Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
2 years ago
Before raw data can be used as scientific evidence, it must be analyzed and summarized. One way of summarizing data is to find t
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6 0
2 years ago
Solve without using a calculator. Substance ab2 is 60.0% a by mass. What is the percent a by mass for substance ab?
Orlov [11]

The percent A by mass for substance AB =<u> 75%</u>

<h3>Further explanation</h3>

Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements

Empirical formula is the mole ratio of compounds forming elements.

From Substance AB₂ is 60.0% A by mass.

Let's say that AB₂ mass = 100 gram, then

mass A = 60 gram

mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram

In compound AB:

Total mass = mass A + mass B

Total mass = 60 + 20 grams = 80 grams

Then the percentage of compound A = (60: 80) = 75%

<h3>Learn more</h3>

Grams of KO₂ needed to form O₂

brainly.com/question/2823257

Keywords : percent mass, substance

#LearnwithBrainly

5 0
2 years ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
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