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3 years ago

El muestre de polvo total, el flujo del muestreo fue de 1.5 lpm, en 8 hrs, Obtener el volumen total de muestreo en m3

Chemistry

1 answer:

fgiga [73]3 years ago

3 0

Answer:

0.72 m3

Explanation:

Primero debes calcular el número de litros que vas a tener en total, para esto conviertes 8 horas a minutos, que son 480 minutos, y lo multiplicas por el flujo de 1.5 litros por minutos, por lo que obtendrás 720 litros. Sabiendo que 1000 litros son 1 m3, podemos calcular que 720 litros son 0.72 m3.

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Write the formulae for the following and calculate the molecular mass for each one of them,

tatuchka [14]

Explanation:

a) NaHCO₃

= 23 + 1 + 12 + 16×3

= 23 + 1 + 12 + 48

= 84 g/mol

b) CaO

= 40 + 16

= 56 g/mol

c) NaCl

= 23 + 35.5

= 58.5 g/mol

8 0

4 years ago

3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?

Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

$\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To balance it, we can simply add another sodium atom on the left. Hence:

$\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

$\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}$

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

$\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}$

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

$\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Given the initial value and the above ratios, this yields:

$\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Cancel like units:

$=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}$

Multiply. Hence:

$=0.3463...\text{ g H$_2$}$

Since we should have two significant values:

$=0.35\text{ g H$_2$}$

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

$\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}$

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

$\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}$

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

$\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}$

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

$\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}$

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: $\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Using the given value and the above ratios, we acquire:

$\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Cancel like units:

$\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}$

Multiply:

$\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}$

Since the resulting value should have three significant figures:

$\displaystyle = 65.2 \text{ g Al$_2$O$_3$}$

So, approximately 65.2 grams of aluminum oxide is produced.

5 0

3 years ago

Read 2 more answers

How is a series circuit different from a parallel circuit?

beks73 [17]

Answer:

C. A series circuit has only one loop and a parallel circuit has two or more loops for the current to flow through

Explanation:

A circuit that are made of one loop is called series circuit. On the other hand, the parallel circuit has at least two loops. The circuit type has nothing to do with open or closed circuit.

If any part of the series circuit got cut, the current will stop flowing since there is only one loop. A parallel circuit has more loop so the circuit might still work even if a part of the circuit got cut.

4 0

3 years ago

A student carefully pipets 10.00ml of ethanol into a beaker. If ethanol has a density of 0.779g/ml, how many grams of ethanol we

lesantik [10]

Answer: 7.79 grams of ethanol were put into the beaker.

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.779 g/mL

Volume of water = 10.00 mL

Putting values in above equation, we get:

$0.779g/mL=\frac{\text{Mass of ethanol}}{10.00mL}\\\\\text{Mass of ethanol}=(0.779g/mL\times 10.00mL)=7.79g$

Thus 7.79 grams of ethanol were put into the beaker.

5 0

3 years ago

if .709 j of heat is added to water and cause the temperature to go up by .036 degrees C what mass of water is present

liberstina [14]

Answer:

0.00471 grams H₂O

Explanation:

To determine the mass, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°C)

-----> ΔT = temperature change (°C)

The specific heat capacity of water is 4182 J/g°C. You can plug the given values into the equation and simplify to isolate "c".

Q = 0.709 J c = 4182 J/g°C

m = ? g ΔT = 0.036 °C

Q = mcΔT <----- Equation

0.709 J = m(4182 J/g°C)(0.036 °C) <----- Insert values

0.709 J = m(150.552) <----- Multiply 4182 and 0.036

0.00471 = m <----- Divide both sides by 150.552

8 0

1 year ago