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2 years ago

What is the number of protons plus the number of neutrons in the nucleus of an atom equivalent to?

Chemistry

1 answer:

Kryger [21]2 years ago

7 0

Answer:

Explanation:

b/c proton + neutron=mass number and

mass number - proton= neutron

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over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th

Aleonysh [2.5K]

Answer : The moles of methane gas could be, $7.90\times 10^{-3}mol$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$

$[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of effusion of fluorine gas

$R_2$ = rate of effusion of methane gas

$n_1$ = moles of fluorine gas = $5.13\times 10^{-3}mol$

$n_2$ = moles of methane gas = ?

$t_1=t_2$ = time = 12.3 min (as per question)

$M_1$ = molar mass of fluorine gas = 38 g/mole

$M_2$ = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

$[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}$

$n_2=7.90\times 10^{-3}mol$

Therefore, the moles of methane gas could be, $7.90\times 10^{-3}mol$

8 0

3 years ago

Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c

jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, $Mg$ = 0.100 mol

Moles of hydrochloric acid, $HCl$ = 0.500 mol

According to the reaction shown below:-

$Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}$

1 mole of Mg reacts with 2 moles of $HCl$

0.100 mol of Mg reacts with 2*0.100 mol of $HCl$

Moles of $HCl$ must react = 0.200 mol

Available moles of $HCl$ = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus, $Mg$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $Mg$ on reaction forms 1 mole of $H_2$

0.100 mole of $Mg$ on reaction forms 0.100 mole of $H_2$

Mole of $H_2$ = 0.100 mol

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K

2.24 L of hydrogen gas, measured at STP, are produced.

4 0

3 years ago

A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

lawyer [7]

$\frac{ V_{1} }{ T_{1} }$ = $\frac{ V_{2} }{ T_{2} }$
$V_{1}$ = 0.0521 L
$T_{1}$ = 293.15 K
$T_{2}$ = 366.65 K

Solve for $V_{2}$

$\frac{0.0521 L}{293.15 K}$ = $\frac { V_{2} }{366.65 K}$

$V_{2}$ = 0.06516 L or 65.2 mL

7 0

3 years ago

A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?

alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = $149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}$

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

$\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}$

The molar mass of gas is 44.19 g/mol

7 0

3 years ago

Classify the processes as endothermic or exothermic.

joja [24]

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

7 0

3 years ago