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3 years ago

Which metal is not reactive with anything

Chemistry

1 answer:

Luden [163]3 years ago

3 0

Answer:Silver, gold, and platinum are metals with the least reactivity.

Explanation:They are found in nature.

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What mass of the protein gelatin is needed to make 0.5 L of a 3 g/L gelatin solution? Show your work.

REY [17]

Answer:

m = 1.5 gram

Explanation:

Given that,

Density of protein gelatin, d = 3 g/L

The volume of protein gelatin, V = 0.5 L

We need to find the mass of the protein gelatin. The density of an object is given by :

d = m/V

Where

m is mass

$m=d\times V\\\\m=3\ g/L\times 0.5\ L\\\\m=1.5\ g$

So, the required mass is 1.5 gram.

6 0

2 years ago

What instrument us normally used to measure atmospheric pressure

Anika [276]

The answer is Barometer.

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3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid. HCl(aq), as described by

melisa1 [442]

Answer:

0.88 g

Explanation:

Using ideal gas equation to calculate the moles of chlorine gas produced as:-

$PV=nRT$

where,

P = pressure of the gas = 805 Torr

V = Volume of the gas = 235 mL = 0.235 L

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{torr}mol^{-1}K^{-1}$

n = number of moles of chlorine gas = ?

Putting values in above equation, we get:

$805torr\times 0.235L=n\times 62.3637\text{ torrHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{805\times 0.235}{62.3637\times 298}=0.01017\ mol$

According to the reaction:-

$MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2$

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.

So,

0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.

Moles of $MnO_2$ = 0.01017 moles

Molar mass of $MnO_2$ = 86.93685 g/mol

So,

$Mass=Moles\times Molar\ mass$

Applying values, we get that:-

$Mass=0.01017moles \times 86.93685\ g/mol=0.88\ g$

<u>0.88 g of $MnO_2(s)$ should be added to excess HCl (aq) to obtain 235 mL of $Cl_2(g)$ at 25 degrees C and 805 Torr.</u>

6 0

3 years ago

How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

Lana71 [14]

Hey...
Use the molarity formula
M=moles/L and then convert to grams

0.07268*0.15=moles
<span>0.010902 mol
</span>Pb(NO3)2
1 mole=331.22g
0.010902 moles=

3.61 g

8 0

3 years ago

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g

luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0

3 years ago