3 years ago

Free pts lollll here ya go

Chemistry

1 answer:

zlopas [31]3 years ago

6 0

Answer:

thanks bro

Explanation:

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Is Kl covalent or ionic?

kotegsom [21]

I believe that the answer is ionic

3 0

3 years ago

A sample of carbon dioxide is contained in a 250.0 mL flask at 0.930 atm and 15.4 °C. How many molecules of gas are in

Yanka [14]

Answer:

We are given:

Volume (V) = 0.25 L

Pressure (P) = 0.93 atm

Temperature (T) = 15.4°C OR 288.4 K

<u>Solving for the number of moles of CO₂:</u>

From the ideal gas equation:

PV = nRT

replacing the variables

0.93 * 0.25 = n (0.082)(288.4)

n = 0.00983 moles

<u>Number of molecules:</u>

Number of moles= 0.00983

number of molecules in 1 mole = 6.022 * 10²³

Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³

Number of molecules = 5.91 * 10²¹

8 0

2 years ago

A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate

saveliy_v [14]

Answer:

$m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

$2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

$m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$