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3 years ago

Amestec 140 g solutie 9% cu 260 g solutie 12%, cu 60 g apa si x g solutie 4%. Ce c procentuala are noua solutie %?

Chemistry

1 answer:

Levart [38]3 years ago

3 0

Answer:

PLEASE SOMEONE SOLVE THE QUESTION OR I'M GONNA FAIL

Explanation:

Purpose and Theory: A tablet of vitamin C contains the active ingredient called ascorbic acid (HC,H,0.).

Ascorbic acid is an organic acid which is a white power and is soluble in water.

In the following microtitration lab you will be grinding up a vitamin C tablet

, and using a 0.10 g sample of the

vitamin C powder, which will be dissolved in 20-40 mL of water. This solution will be used to determine the

quantity of ascorbic acid in a commercial vitamin C tablet.

Groups

Lab 2

# of Drops 1 or Drops 2 of Drops 3

Hadia

Data & Analysis:

Complete the missing information in the following data table [24]

Trial

Volume of NaOH (drops)

0 05mL drop

(NaOH (O 0050mol/L)

Volume of NaOH (ML)

Average Volume of NaOH (ml)

T I

O 0050

00050

The balanced chemical reaction between ascorbic acid and sodium hydroxide is:

HC,H,O + NaOH NaC,H,Oye H,O,

1. Stoichiometrically calculate the mass (in grams) of ascorbic acid in the powdered sample.

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Calculate the energy required to produce 6.0 moles of Cl2O7 using the following reaction 2 Cl2 + 7 O2 + 130kcal -> 2 Cl207

PilotLPTM [1.2K]

Answer:

455 Kcal

Explanation:

2Cl2(g) + 7O2(g) + 130kcal → 2Cl2O7(g)

Rearranging we get,

2Cl2(g) + 7O2(g) → 2Cl2O7(g) Δ H = 130 kcal . mol⁻¹

So for per mol reaction will be as above.

In case of 7 mols of product, we need 7/2 mole ratio x 130 = 455 Kcal

5 0

2 years ago

Although each famous crater looks different, they have the same basic circular shape. Visit the Rio Cuarto craters in Argentina

Bumek [7]

Answer:

The controversy behind this craters is that thanks to the angle of the object's impact they could be habitable. Also, they could study them more.

Explanation:

The reason behind this answer is that majority of craters have an angle of impact of aorund45º. While the craters in the province of Cordoba, Argentina have an angle of impact of 15º. MAking them really plain, so they could be habitable for several species. Also, the materials found in the place suggest nothing more than quartz, so they represent no danger to the inhabitants, humans, or not. But more importantly that they have only received one major research study. Instead of a big variety, and they could be studied with more detail.

6 0

3 years ago

Pleas help with 2 and 4 for brainliest

Snezhnost [94]

mass of pentane : = 30.303 g

moles of Al₂(CO₃)₃ : = 0.147

<h3>Further explanation</h3>

Given

1. Reaction

C₅H₁₂+8O₂→6H₂O+5CO₂.

45.3 g water

2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂

37.2 MgCO₃

Required

mass of pentane

moles of Al₂(CO₃)₃

Solution

1. mol water = 45.3 : 18 g/mol = 2.52

From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :

= 1/6 x mol H₂O

= 1/6 x 2.52

= 0.42

Mass pentane :

= mol x MW

= 0.42 x 72.15 g/mol

= 30.303 g

2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44

mol Al₂(CO₃)₃ :

= 1/3 x mol MgCO₃

= 1/3 x 0.44

= 0.147

4 0

2 years ago

All atoms of the same elements have the same what?

Flura [38]

Answer:

The same number of proteins in their nucleus.

4 0

2 years ago

Read 2 more answers

a 4.50 g coin of copper absorbed 54 calories of heat. what was the final temperature of the copper if the initial temperature wa

vlada-n [284]

Answer:

Final temperature = T₂ = 155.43 °C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of coin = 4.50 g

Heat absorbed = 54 cal

Initial temperature = 25 °C

Specific heat of copper = 0.092 cal/g °C

Final temperature = ?

Solution:

Q = m.c. ΔT

ΔT = T₂ -T₁

Q = m.c. T₂ -T₁

54 cal = 4.50 g × 0.092 cal/g °C × T₂ -25 °C

54 cal = 0.414 cal/ °C × T₂ -25 °C

54 cal /0.414 cal/ °C = T₂ -25 °C

130.43 °C = T₂ -25 °C

130.43 °C + 25 °C = T₂

155.43 °C = T₂

4 0

3 years ago