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3 years ago

What are the only elements that exist in nature as uncombined

Chemistry

1 answer:

mario62 [17]3 years ago

5 0

Answer:

The noble gases (Argon, Neon, Xenon, Helium, Krypton, Radon)

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Need help plz asap .......

olganol [36]

Answer:

Purple flowers

Explanation:

Usually the dominant allele is a capital letter.

From the question, purple flowers are dominant to white flowers while white are recessive.

PP = Purple

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Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

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3 years ago

Write a nuclear reaction for the neutron-induced fission of u?235 to produce te?137 and zr?97. how many neutrons are produced in

alexandr1967 [171]

This is a neutron induced fission, therefore a neutron will be added to the U²³⁵ to cause the reaction, and thus it will be added to the left side. There will be unknown number of neutrons produced and thus we put this on the right hand side.
n₁ + U²³⁵ = Te¹³⁷ + Zr ⁹⁷ + xn1 ( n1 to mean a neutron of mass 1)
To balance the masses on both sides of the equation;
1 + 235 = 137 +97+ x
x = 2
the end reaction will be
n₁ +U²³⁵ = Te¹³⁷ + Zr⁹⁷ + 2 n₁

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