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2 years ago

If a tank of gas contains 4 L of N O2 how many molecules are in it

Chemistry

1 answer:

Jobisdone [24]2 years ago

3 0

but heres a way to solve it

An athlete takes a deep breath, inhaling 1.85 L of air at 21°C and 754 mm Hg.

How many moles of air are in the breath? How many molecules?

Gas constant, R= 8.314 J mol ¹ K-1

PV = nRT

= 0.08206 L atm mol-1 K-1

= 62.36 L Torr mol-1 K-1 -

1 atm = 760 mm Hg = 760 Torr

754 Forr 1.85€

6236 Jerr 294K

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What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

$Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+$

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

Now put all the given values in this expression, we get:

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

The concentration of $H^+$ = x = 0.00212 M

Now we have to calculate the pH of solution.

$pH=-\log [H^+]$

$pH=-\log (0.00212)$

$pH=2.67$

Therefore, the pH of the solution is, 2.67

8 0

3 years ago

Compute 214.056 + 9.3456. Round the answer appropriately

shusha [124]

214.0560
+ 9.3456
=223.4016

3 0

3 years ago

Helppppp with this pleasee

vazorg [7]

Answer:

A. 1, 2, 5

Explanation:

Count the number of Ns in the formula.

- Hope that helped! Please let me know if you need a further explanation.

3 0

3 years ago

44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?

postnew [5]

Answer:

112 mL

Explanation:

The formula for percent by volume is

$\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%$

If you have 250 mL of a solution that is 44.8 % v/v,

$\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}$

7 0

3 years ago

A small hole in the wing of a space shuttle requires a 20.7-cm² patch.

cricket20 [7]

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or, 1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 × 10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

#SPJ4

8 0

1 year ago

If a tank of gas contains 4 L of N O2 how many molecules are in it￼

If a tank of gas contains 4 L of N O2 how many molecules are in it