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3 years ago

What are some problems an astronaut might experience while traveling in space?

Chemistry

1 answer:

Jlenok [28]3 years ago

7 0

Answer:

The environment of space is lethal without appropriate protection: the greatest threat in the vacuum of space derives from the lack of oxygen and pressure, although temperature and radiation also pose risks. The effects of space exposure can result in ebullism, hypoxia, hypocapnia, and decompression sickness.

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What is the most appropriate way to present the data below?

Arada [10]

Answer:

I think bar graph is the appropriate way to present the data.

4 0

3 years ago

Read 3 more answers

CH2OHCH2OH is a general example of: an ethyl alcohol a methyl alcohol a polyhydroxyl alcohol an organic acid

RoseWind [281]

CH2OHCH2OH is a general example of a polyhydroxyl alcohol. A polyhydroxyl alchol is one in which there are two hydroxyl groups present in the substance. The –OH group attached to both the carbon atoms.

4 0

3 years ago

Read 2 more answers

en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es

Alinara [238K]

The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.

solution 1

composition :

35% acid :

$\tt 0.35\times 800~g=280~g$

water :

$\tt 800-280=520~g$

solution 2(new solution)

composition :

water

$\tt 520-(80~ml\times 1~g/ml)=440~g$

Total mass of new solution after water evaporated

$\tt 280(acid)+440(water)=720~g$

%mass of acid in a new solution

$\tt \dfrac{280}{720}\times 100\%=38.9\%$

5 0

2 years ago

Suppose that 25.0 mL of 0.440 M sodium chloride is added to 25.0 mL of 0.320 M silver nitrate. How many moles of silver chloride

d1i1m1o1n [39]

The number of moles of silver chloride that will precipitate is 0.008 mole

From the question,

We are to determine the number of moles of silver chloride that will precipitate

First,

We will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

NaCl + AgNO₃ → AgCl + NaNO₃

This means

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride and 1 mole of sodium nitrate

Now, we will determine the number of moles of each reactant present

For sodium chloride (NaCl)

Concentration = 0.440 M

Volume = 25.0 mL = 0.025 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of NaCl present = 0.440 × 0.025

Number of moles of NaCl present = 0.011 mole

For silver nitrate (NaNO₃)

Concentration = 0.320 M

Volume = 25.0 mL = 0.025 L

∴ Number of moles of NaNO₃ present = 0.320 × 0.025

Number of moles of NaNO₃ present = 0.008 mole

From the balanced chemical equation,

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride

Then,

0.008 mole of sodium chloride will react with the 0.008 mole of silver nitrate to produce 0.008 mole of silver chloride

∴ 0.008 mole of silver chloride will be produced

Hence, the number of moles of silver chloride that will precipitate is 0.008 mole

Learn more here: brainly.com/question/18434602

4 0

2 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago