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3 years ago

Explain this

Chemistry

2 answers:

Alenkasestr [34]3 years ago

3 0

Answer:

Because Argon is a inert gas

Explanation:

Inert gases don't take place in any experiments

Kay [80]3 years ago

3 0

Answer:

Below.

Explanation:

The sodium reacts strongly with the chlorine to form NaCl and the heat of reaction continues the brightness, whereas it does not react with the inert gas argon and soon cools down.

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The ksp of manganese(ii) carbonate, mnco3, is 2.42 × 10-11. calculate the solubility of this compound in g/l.

Vikki [24]

Answer : The solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

Solution : Given,

$K_{sp}=2.42\times 10^{-11}$

Molar mass of $MnCO_3$ = 114.945g/mole

The balanced equilibrium reaction is,

$MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3$

At equilibrium s s

The expression for solubility constant is,

$K_{sp}=[Mn^{2+}][CO^{2-}_3]$

Now put the given values in this expression, we get

$2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L$

The value of 's' is the molar concentration of manganese ion and carbonate ion.

Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.

$s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L$

Therefore, the solubility of this compound in g/L is $565.414\times 10^{-6}g/L$ .

8 0

3 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

Help please please please please

Rom4ik [11]

Answer:

D. 18.7 grams

Explanation:

The coefficients are the key.

Create a proportion with them, and the molar mass, and then solve for x:

$\frac{2}{3} = \frac{x}{28} \\ 3x = 2(28) \\ 3x = 56 \\ x = 18.666... = 18.7$

7 0

3 years ago

A b c d e f g h i j k l m n o p q r s t u v w x y z

enyata [817]

It’s B. Substitution hope this helps

6 0

2 years ago

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Of the following substances, only __________ has London dispersion forces as its only intermolecular force. CH3OH NH3 H2S CH4 HC

QveST [7]

Answer:

CH₄

Explanation:

CH₃OH has hydrogen bonding due to the OH group present

NH₃ also has hydrogen bonding due to the NH bonds

H₂S has dipole-dipole forces present due to the polar SH bonds

HCl also has dipole-dipole forces due to the polar HCl bond

7 0

3 years ago