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andrey2020 [161]

3 years ago

12

Two hydrogen nuclei repel one another due to the positive charges of the protons within them. At some separation distance this r

epulsion is overcome and the hydrogen nuclei can fuse to form helium. What force allows two protons to "stick" together? A) gravity B) electromagnetism C) weak nuclear force D) strong nuclear force

2 answers:

ValentinkaMS [17]3 years ago

5 0

Answer: The answer to your question is D) Strong nuclear force

Explanation:

In nature, there are four fundamental interactions: electromagnetism, gravitation, weak nuclear force and strong nuclear force.

The Strong nuclear force is responsible for binding protons and neutrons to form a new nucleus.

Weak nuclear force is responsible for radioactive decay which means splitting nucleus.

Vesna [10]3 years ago

3 0

Answer:

D (strong nuclear force)

Explanation:

got the answer correct on USA test prep

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What is the difference between physical and chemical change in chemistry?

VARVARA [1.3K]

physical changes are those changes which can be reversed in their previous form like ice to water

chemical changes are those changes which cannot be reversed in their previous form

5 0

3 years ago

Read 2 more answers

When sodium metal is added to water, the following reaction occurs:

zheka24 [161]

Answer:

d. n H2(g) = 0.034 mol

Explanation:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

2 - Na - 2

4 - H - 4

2 - O - 2

∴ n Na(s) = 0.068 mol

⇒ n H2(g) = ( 0.068 mol Na(s) )( mol H2(g) / 2 mol Na(s) )

⇒ n H2(g) = 0.034 mol

8 0

3 years ago

g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h

Setler79 [48]

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

3 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]

<u>Answer:</u> The mass of methane burned is 12.4 grams.

<u>Explanation:</u>

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.

<u>For process 1:</u>

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

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:

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

<u>For process 2:</u>

$q_2=m\times L_v$

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:

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

<u>For process 3:</u>

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

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

2-Ethoxy-2,3-dimethylbutane reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction

Leno4ka [110]

Answer:

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

Explanation:

2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.

On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.

But ethanol reacts further with conc HI to give iodoethane.

Therefore,

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

This is all shown in the attached image.

Hope this Helps!!!

6 0

4 years ago