What's an example of using evaporation to separate a mixture in everyday life?

dlinn [17]

Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693. $\overline{3168}$ grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

$\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}$

$\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g$

The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

$\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}$

$\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g$

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693. $\overline{3168}$ grams

6 0

3 years ago

(d) If you have 9.44 g of NazPO4, how many grams of AgNO, will be needed for complete

AlladinOne [14]

30.6 g of AgNO₃

Explanation:

We have the following chemical reaction:

Na₃PO₄ + 3 AgNO₃ → Ag₃PO₄ + 3 NaNO₃

Now we calculate the number of moles of Na₃PO₄:

number of moles = mass / molecular weight

number of moles of Na₃PO₄ = 9.44 / 164 = 0.06 moles

Now we formulate the following reasoning:

if 1 mole of Na₃PO₄ will react with 3 moles of AgNO₃

then 0.06 moles of Na₃PO₄ will react with X moles of AgNO₃

X = (0.06 × 3) / 1 = 0.18 moles of AgNO₃

mass = number of moles × molecular weight

mass of AgNO₃ = 0.18 × 170 = 30.6 g

Learn more about:

mole concept

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3 0

3 years ago

For the reaction 2 A + 4 B → 2 C + 4 D, what is the absolute magnitude of the rate of change for [B] when [C] is increasing at 4

Hoochie [10]

Answer:

$r_B=9.6M/s$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to relate the rate of formation of C and the rate of consumption of B as shown below:

$\frac{r_B}{-4}=\frac{r_C}{2} \\\\$

Thus, we solve for the rate of change of B as shown below:

$r_B=\frac{-4r_C}{2}\\\\r_B=\frac{-4(4.8M/s)}{2} \\\\r_B=9.6M/s$

Best regards!

8 0

3 years ago

HELPPPPPPPPPPPPP meeeeeeeee

Jobisdone [24]

Answer:

100%

Explanation:

If all labels in the square are “t” then there is no way that the plants will not be tall. Therefore there is a 100% chance the plant will be tall.

8 0

3 years ago

Read 2 more answers

When a 0.680 g sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 370 g of water fro

Lena [83]

Answer:

the energy value of the olive oil kJ/g = 36.880 KJ/g

Explanation:

we can calculate energy gained by the water which generated by the reaction from this formula :

Q = mc∆t

where:

m is the mass of water= 370 g

c specific heat of water = 4.184 J/g·°C

∆t is the change or difference in temperature in °C = 38.9 - 22.7 = 16.2 °C

by substitution:

Q = 370g * 4.184 J/g.°C* 16.2°C

= 25078.896 Joule

the energy value of the olive oil = Q / mass of olive oil

= 25078.896 Joule/ 0.68 g

=36880 J/g

= 36880 / 1000 = 36.880 KJ/g

6 0

3 years ago