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3 years ago

Which one of the following substances is a solution?

Chemistry

2 answers:

Nostrana [21]3 years ago

7 0

Hdhdhdjdndjsjshshshhd

Sphinxa [80]3 years ago

4 0

A . Clay this solution is used in old buildings

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How many moles of carbon do you have in a trillion carbon

Dmitry_Shevchenko [17]

Answer:

$1.7\times 10^{-12}$ moles of carbon

Explanation:

1 trillion = $10^{12}$

Avogadro number :It is the number of particles that are present in 1 mole of substance.

$N_{0} = 6.022\times 10^{23}$

So,

$6.022\times 10^{23}$ atoms = 1 mole carbon

1 atom =

$\frac{1}{6.022\times 10^{23}}$ mole

1 trillion atom =

$\frac{1}{6.022\times 10^{23}}\times 10^{12}$ mole

$1.660\times 10^{-12}$ mol

Two significant figure

$1.7\times 10^{-12}$ moles of carbon

4 0

3 years ago

How does water heat Earth?

nata0808 [166]

Answer:

I hope it's helpful for you...

Explanation:

water traps heat and holds it deep inside the ocean .....

I think it's correct answer

5 0

3 years ago

Read 2 more answers

How many molecules are in 25g of Na2SO4? POINTS!! Please help! Im so frustrated.

erik [133]

Hey there!

Molar mass Na2SO4 = <span>142.04 g/mol

Number of moles:

n = m / mm

n = 25 / 142.04

n = 0.176 moles of Na2SO4

Therefore, </span>use the Avogadro constant

1 mole Na2SO4 ------------------- 6.02x10²³ molecules
0.176 moles Na2SO4 ------------ molecules ??

0.176 x ( 6.02x10²³ ) / 1

=> 1.059x10²³ molecules of Na2SO4

hope this helps!

6 0

4 years ago

Which material has been recycled for over 3,000 years?

Juliette [100K]

Answer:

B. glass

Explanation:

I did research about the topic.

4 0

3 years ago

Read 2 more answers

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

4 years ago