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3 years ago

What is the concept of galactic recycling?

1 answer:

5 0

Galactic recycling is a natural phenomenon in which the stars expel some gas into the space so that it would mix with the interstellar medium. As a result, this would produce new younger stars. So, basically, the concept done here is the mass and energy conservation. In order to create new species, the energy must come from another source.

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A 53. 5 mL ample of an 5. 4 % (m / v) KBr olution i diluted with water o that the final volume i 205. 0 mL Expre your anwer to t

gogolik [260]

The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

$C_{1}$ $V_{1}$ = $C_{2}$ $V_{2}$ , where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

Keep the values to calculate final concentration.

$C_{2}$ = (53.5 × 5.4)/205.0

Performing multiplication on Right and Side

$C_{2}$ = 288.9/205.0

Performing division on Right Hand Side

$C_{2}$ = 1.4%

Hence, the final concentration is 1.4%.

Learn more about concentration -

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The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

3 0

1 year ago

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

erica [24]

Answer: The amount of $P_4O_{10}$ formed is 469.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

8 0

2 years ago

Products for combustion

V125BC [204]

Answer:

Carbon Dioxide

Carbon Monoxide

Sulfur Dioxide

Nitrogen Oxides

Lead

Particulate Matter

Explanation:

7 0

1 year ago

Read 2 more answers

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 381 mL . Express yo

BaLLatris [955]

Multiply the volume by the density. 1 ml = 1 cm^3

1.11 g/cm^3 * 1 cm^3/mL * 386 mL = 428 g

All your units cancel out except for g which is what you would expect since you want units of mass.

3 0

2 years ago

Read 2 more answers

Sodium bicarbonate is sometimes used to neutralize acid spills. If 2.0 L of 1.0 M H2SO4 is spilled, what mass of sodium bicarbon

UkoKoshka [18]

Answer:

168.0 g

Explanation:

First thing, write a balanced chemical equation:

$H_{2}SO_{4}+ 2NaHCO_{3} ----> Na_{2}SO_{4} + 2H_{2}O+2CO_{2}$

n(H2SO4) = concentration * volume

= 1.0 M * 2.0 L

= 2.0 mol

According to the balanced equation, 1 mol of acid requires 2 mol of sodium bicarbonate. This means that 2 mol of acid requires 2 mol of sodium bicarbonate. What mass of sodium bicarbonate is this?

mass (NaHCO3) = number of moles * molar mass

= 2.0 mol * 84.0065 g/mol

= 168.0 g

4 0

3 years ago