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NISA [10]
3 years ago
6

what volume of carbon 4 oxide measured at STP will produce when 21 gram of sodium hydrogen carbonate for is completely decompose

d accor. ding to the equation. (Na=23. H=1 C=12 O=16) molar vol =22.4dmcube. pleeeeeeeeeeese help​,​
Chemistry
1 answer:
Arlecino [84]3 years ago
8 0

The volume of CO₂ at STP = 2.8 L

<h3>Further explanation</h3>

Given

21 gram of sodium hydrogen carbonate-NaHCO₃

Required

Volume of CO₂

Solution

The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :

<em> 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)</em>

mol of NaHCO₃ :

= mass : MW NaHCO₃

= 21 g : 84 g/mol

= 0.25

From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.25

= 0.125

At STP, 1 mol gas = 22.4 L, so for 0.125 mol :

= 0.125 x 22.4 L

= 2.8 L

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3.25 x 10+8 nm2 divide by 6.5 x 10+6 nm =
Vesnalui [34]

Answer:  50 nm

Explanation:  Two steps:

1.  Divide 3.25/6.5 = 0.5

2.  Divide 10^8/10^6 = 10^2

nm^2/nm = nm

Combine:  0.5x10^2 nm

or 50 nm

8 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
How many grams are in CO2?
Sati [7]
There are 44grams in CO2
5 0
3 years ago
Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C,
AlladinOne [14]

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

7 0
3 years ago
What was the result of heating the mixture? All BUT ONE choice is correct.
snow_tiger [21]

Answer:

w gang alright

Explanation:

ay its b alright

4 0
3 years ago
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