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3 years ago

5

How long does it take for a life saver hard candy in hot water to dissolve

1 answer:

madam [21]3 years ago

7 0

Answer:

around a minute

Explanation:

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I NEED AN ANSWER ASAP!!! HELP ME PLZ

RSB [31]

Answer:

No2

Explanation:

4 0

3 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

An unbalanced chemical equation is shown:

ExtremeBDS [4]

There are 2 Ca, 4 H (2*2) and 4 O (2+2) on the left hand side. So there should be the same number of atoms 2Ca and 4(O+H) on the right hand side. The reaction is a synthesis (making something new), rather than a decomposition (destroying something).

The answer is Two molecules of Ca(OH)2 should be produced during the synthesis reaction.

4 0

4 years ago

Read 2 more answers

New oceanic lithosphere forms as a result of

Oksana_A [137]

Answer:

A. sea-floor spreading.

Explanation:

The sea-floor spreading is the result of the new oceanic lithosphere

3 0

3 years ago

Read 2 more answers

Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M aceto

GarryVolchara [31]

Answer:

(a) $m=2.69m$

(b) $x_{LiBr}=0.099$

(c) $\% LiBr=18.9\%$

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

$m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g$

Next, we compute the mass of the solution:

$m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g$

Then, the mass of the solvent (acetonitrile) in kg:

$m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg$

Finally, the molality:

$m=\frac{1.80mol}{0.670kg} \\\\m=2.69m$

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

$n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol$

Then, the mole fraction of lithium bromide:

$x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099$

(c) Finally, the mass percentage with the previously computed masses:

$\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%$

Regards.

4 0

4 years ago