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timama [110]

4x = -12

First, divide both sides by 4. / Your problem should look like: $x = -\frac{12}{4}$
Second, since 4 goes into 3 to get 12, simplify it by 3. / Your problem should look like: $x = -3$

Answer: x = -3

6 0

3 years ago

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A 15.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.0 g of He and 4.14 g of O2 at 298 K. Calculat

EleoNora [17]

Answer: The mole fraction of Helium is 0.978 and that of oxygen gas is 0.022

Explanation:

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

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

For helium:

Given mass of helium = 23 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{23g}{4g/mol}=5.75mol$

For oxygen gas:

Given mass of oxygen gas = 4.14 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{4.14g}{32g/mol}=0.129mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For helium:

Moles of helium = 5.75 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

$\chi_{(He)}=\frac{5.75}{5.879}=0.978$

For oxygen gas:

Moles of oxygen gas = 0.129 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

$\chi_{(O_2)}=\frac{0.129}{5.879}=0.022$

Hence, the mole fraction of Helium is 0.978 and that of oxygen gas is 0.022

3 0

3 years ago

I need help with two questions.

vodomira [7]

Answer:

1) their atomic numbers

2) periodic table

8 0

3 years ago

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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

Question 5

pantera1 [17]

Answer: The coefficient for the diatomic oxygen (O2) is 3.

Explanation:

To know the coefficient for the diatomic Oxygen, we need to balance the equation.

Fe + O2 -------> Fe2O3

LHS of the equation; Fe = 1 , O2 = 1

RHS of the equation; Fe = 2 , O = 3

∴ Multiply 'Fe' on the LHS of the equation by 4 and O2 by 3

Doing that will give the balance equation which is;

4 Fe + 3 O2 --------> 2 Fe2O3

The coefficient for the diatomic oxygen (O2) as seen from the equation is 3.

7 0

3 years ago