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Roman55 [17]
3 years ago
14

How many moles of sodium bicarbonate are in 1.50 kg of the compound?

Chemistry
1 answer:
salantis [7]3 years ago
3 0

Answer:

n=17.85 moles

Explanation:

Given mass is, m = 1.5 kg = 1500 g

The molar mass of sodium bicarbonate is, M = 84.007 g/mol

We need to find the no of moles in 1.5 kg of Sodium bicarbonate . We know that, no of moles is equal to given mass divided by molar mass.

n=\dfrac{m}{M}\\\\n=\dfrac{1500}{84.007 }\\\\n=17.85

<h2>So, there are 17.85 moles in 1.5 kg of Sodium bicarbonate.</h2>
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storchak [24]

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Rift Valley Facts/Characteristics

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Hope this helps.

5 0
3 years ago
Draw the following isomers of the alkanes of molecular formula C-H16
Andre45 [30]

I don't know how to draw it but I can solve it

7 0
3 years ago
How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?
Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g)  + 5O2(g)  ----> 3CO2(g)   + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0


7 0
3 years ago
Read 2 more answers
igneous rock that cools very quickly when it forms may have a texture that is______and______with no visible____
Fudgin [204]
<span>Igneous rock that cools very quickly when it forms may have a texture that is smooth and shiny with no visible grain.</span>
6 0
3 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
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