Help for brainlist and extra points!!

A student increases the temperature of a 417cm³ balloon from 278k to 231k. Assuming constant pressure, what should the new volum

viva [34]

Charles law states that volume of gas is directly proportional to temperature at constant pressure
V/T = k
where V - volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation

in the question it states that the temperature has been increased from 278 K to 231 K but it should actually be temperature is decreased from 278 K to 308 K
substituting the values in the equation
$\frac{417cm^{3} }{278K} = \frac{V}{308 K}$
V = 462 cm³
the answer should be D. 462 cm³

3 0

3 years ago

The process of pouring a mixture through a filter to catch the larger particles, while allowing the smaller particles to pass th

ki77a [65]

It’s is b. Filtration

7 0

3 years ago

When you heat a substance what happens to the molecules

tatyana61 [14]

I'm pretty sure they come closer, and on the small chance i'm wrong, they separate.<span />

6 0

3 years ago

Read 2 more answers

How many liters of oxygen gas, at standard

Karo-lina-s [1.5K]

Answer:

Explanation:

For the balanced reaction:

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0

3 years ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s