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3 years ago

What is the charge of a beta particle? 0 1– 2+ 4+

Chemistry

2 answers:

marusya05 [52]3 years ago

8 0

Its 2+! Dont mind this hdhdhdjhdndbdbdbdbsnnsjdndsnjd

Vlada [557]3 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

A beta particle is a particle with a negatively charged electron. Symbol of a beta particle is $^{0}_{-1}\beta$ .

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.

Symbol of an alpha particle is $^{4}_{2}\alpha$ .

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is $^{0}_{0}\gamma$ .

Therefore, we can conclude that the charge of a beta particle is 1-.

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AVprozaik [17]

Answer:

If a substance can be dissolved it is called soluble.

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8 0

3 years ago

Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago

A balloon is filled with 3.0 l of helium at 20oc. at what temperature will the balloon be one-half that volume (1.5 l

timama [110]

Answer1. A certain mass of gas in a 2.00 L container has a pressure of 164 kPa. Calculate the new pressure of the gas if the volume of the container is reduced to 1.00L

7 0

3 years ago

Calculate the percent dissociation of trimethylacetic acid(C6H5CO2H) in a aqueous solution of the stuff.

Sophie [7]

Answer:

1.112%.

Explanation:

Step one : write out the correct acid dissociation reaction. This is done below:

C6H5CO2H <=====> C6H5CO2^- + H^+.

At equilibrium, the concentration of C6H5CO2H = 0.5 M - x and the concentration of C6H5CO2^- = x and the concentration of H^+ = x.

The ka for C6H5CO2H = 6.3 × 10^-5.

Step two: find the value for the concentration of C6H5CO2^- and H^+. This can be done by using the equilibrium dissociation Constant formula below;

Ka = [C6H5CO2^- ] [H^+] / C6H5CO2H.

ka = [x] [x] / [0.5 - x].

6.3 × 10^-5 = (x)^2 / [0.5 - x].

x^2 = 3.15 ×10^-5 - 6.3 × 10^-5 x.

x^-2 + 6.3 × 10^-5 x - 3.15 × 10^-5.

Solving for x we have x= 0.0055632289702004 = 0.00556.

Therefore, at equilibrium the concentration of H^+ = 0.00556 M and the concentration of C6H5CO2H= O.5 - 0.00556 = 0.494 M.

Step three: Calculate the equilibrium pH. This stage can be ignored for this question since we are not asked to calculate for the pH.

The formular for pH = - log [H^+].

pH= - log [0.00556].

pH= 2.255.

Step four: find the percentage dissociation.

Percentage dissociation = ( [H^+] at equilibrium/ [ C6H5CO2H] at Initial concentration ) × 100%.

Percentage dissociation=( 0.00556/ 0.5 ) × 100.

=Percentage dissociation= 1.112%.

3 0

3 years ago

What are the different characteristics that can be used to identify an unknown mineral

brilliants [131]

These are just a couple ways to identify unknown minerals,touch taste smell

4 0

4 years ago

Read 2 more answers