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padilas [110]
3 years ago
5

Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo

l/s) (ii) Mass flow rate of O (Mw=16.00) in this stream (8 0/s)? (iii) Mass flow rate of H (Mw=1.00) in this stream (g H/s)?
Chemistry
1 answer:
Arada [10]3 years ago
3 0

Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

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Use the following balanced equation to answer the questions below.
Blababa [14]

Answer:

A. 4.5 mol Mg(OH)₂

B. 6 mol NaOH

Explanation:

Let's consider the following balanced equation.

Mg(NO₃)₂ + 2 NaOH ⇒ Mg(OH)₂ + 2 NaNO₃

PART A

The molar ratio of NaOH to Mg(OH)₂ is 2:1. The moles of Mg(OH)₂ produced from 9 moles of NaOH are:

9 mol NaOH × 1 mol Mg(OH)₂/2 mol NaOH = 4.5 mol Mg(OH)₂

PART B

The molar ratio of NaOH to NaNO₃ is 2:2. The moles of NaOH needed to produce 6 moles of NaNO₃ are:

6 mol NaNO₃ × 2 mol NaOH/2 mol NaNO₃ = 6 mol NaOH

5 0
2 years ago
2Na2SO4 :How many atoms for each element are in the formula?<br><br> S=<br> O=<br> Na=
yuradex [85]

Answer:

S=  2(1)  = 2

O=  2(4)  = 8

Na= 2(2)  = 4

Explanation:

The given compound is:

          2Na₂SO₄  

An element is a distinct substance that cannot be split up into simpler substances.

So;

  Number of atoms of elements here are:

S=  2(1)  = 2

O=  2(4)  = 8

Na= 2(2)  = 4

6 0
2 years ago
Please Help !! <br><br> The weak base ionization<br><br> constant (Kb) for CIO is<br><br> equal to:
Zolol [24]

Answer:

k it I did I'd help ya

Explanation:

sirry

3 0
3 years ago
299792458 m/s in scientific notation
erastova [34]
2.998e^8 is how I would write it. If you want it with the least amount of decimals, use 3e^8
6 0
3 years ago
Read 2 more answers
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
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