Does this make a reaction? Because both products are soluble.

Which of the following is NOT a sign of a chemical reaction:

Sergeeva-Olga [200]

The answer is B. Precipitation

7 0

3 years ago

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{31.862\ g}{63.546\ g/mol}$

Moles of copper = 0.5014 moles

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 284\ g$ = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{193.12\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.0649 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.0028 moles

Given:

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 25.76 L

4 0

3 years ago

Calculate the amount of heat needed to boil 159.g of water

MissTica

Okay I did the math and I'm guessing around 18*C

5 0

3 years ago

Consider the following reaction: C6H6 + O2 \longrightarrow ⟶ CO2 + H2O 39.7 grams of C6H6 are allowed to react with 105.7 g of O

Ivenika [448]

Answer:

116.3 grCO2

Explanation:

1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side

C6H6 +15/2 O2⟶ 6CO2 +3 H2O

2nd - we calculate the limiting reagent

39.2gr C6H6*(240grO2/78grC6H6)=120 grO2

we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent

3rd - we use the limiting reagent to calculate the amount of CO2 in grams

105.7grO2*(264grCO2/240grO2)=116.3 grCO2

7 0

3 years ago

What three facts made Pluto difficult to find?

BigorU [14]

1 Its low gravitational influence.
2 Its extremely faint, compared to all the other objects out there.
3 Its Size

6 0

3 years ago

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