All categories

2 years ago

What do ALL states of matter have in common?

Chemistry

1 answer:

VLD [36.1K]2 years ago

3 0

Answer:

The common thing among the three states of matter is - They are made up of small tiny particles. They have a particular mass and can occupy space . This three states have volume in it . The atoms of this three states have force of attractions between them .

Explanation:

hope this helps <3

You might be interested in

Describe why the periodic table needed to be reordered in 1913.

Norma-Jean [14]

Answer:

1913. Henry Moseley determined the atomic number of each of the known elements. He realized that, if the elements were arranged in order of increasing atomic number rather than atomic weight, they gave a better fit within the 'periodic table'.

Explanation:

GOO

7 0

2 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Write a net ionic equation that describes a slightly basic solution resulting from dissolving nac2h3o2 in water. Identify all of

Natali [406]

Answer: The net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

Explanation:

The chemical equation for the given reaction is as follows.

$CH_{3}COONa(aq) + H_{2}O(l) \rightleftarrow CH_{3}COOH(aq) + NaOH(aq)$

We know that a strong acid or base will dissociate completely into a solvent whereas a weak acid or base dissociates partially into the solvent. Hence, the ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + Na^{+}(aq) H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + Na^{+}(aq) + OH^{-}(aq)$ Now, we will cancel the spectator ions from the above equation. Therefore, the net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

or, $C_{2}H_{3}O^{-}_{2}(aq) + H_{2}O(l) \rightleftharpoons HC_{2}H_{3}O_{2}(aq) + OH^{-}(aq)$

4 0

3 years ago

A gas occupies a volume of 180 mL at 35oC and 740 mmHg. What is the volume of the gas at STP?

Tasya [4]

Answer:

there you go

Explanation:

6 0

3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago