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mihalych1998 [28]

3 years ago

14

Which elements have isotopes with the same number of protons and neutrons

1 answer:

yan [13]3 years ago

7 0

Answer:

Atoms of the same element, containing the same number of protons, but different numbers of neutrons, are known as isotopes. Isotopes of any given element all contain the same number of protons, so they have the same atomic number

Explanation:

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Consider the reaction mg(oh)2(s)→mgo(s)+h2o(l) with enthalpy of reaction δhrxn∘=37.5kj/mol what is the enthalpy of formation of

Anni [7]

Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = <span>−924,5 kJ/mol.
</span>ΔHf(H₂O) = <span>−285,8 kJ/mol.
</span>ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
<span>ΔHrxn=∑productsΔHf−∑reactantsΔHf.
</span>ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.

7 0

3 years ago

Read 2 more answers

What is the name for CH 3 CH 2 COCHCH 3 CH(CH 3 ) 2 ?

Paraphin [41]

Answer:

Butanoic acid

Explanation:

The IUPAC name of CH3CH2CH2COOH is:

The IUPAC name for a given compound is Butanoic acid.

4 0

2 years ago

The equation below shows the products formed when a solution of silver nitrate (AgNO3) reacts with a solution of sodium chloride

d1i1m1o1n [39]

I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant

3 0

3 years ago

Read 2 more answers

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

I need somebody to help me with my homework pls i dont get it .

zubka84 [21]

26. B
27. D
28. C
Hope this helped ☺️

7 0

3 years ago