Please help me which one

Find the formula for the hydrate<br>0.737 g MgSO3 and 0.763 g H2O

babunello [35]

The required formula of hydrate is MgSO₃.6H₂O.

<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
n = W/M, where

W = given mass
M = molar mass

Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol

Moles of H₂O = 0.763g / 18g/mol = 0.04 mol

Number of H₂O molecule = 0.04/0.007 = 5.7 = 6

So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

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6 0

1 year ago

What is bed rock made of

Darya [45]

Answer:

Bedrock is the hard, solid rock beneath surface materials such as soil and gravel.

Bedrock can be made of most types of rock, such as granite, limestone, or sandstone.

Hope this helps!

8 0

3 years ago

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The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con

bonufazy [111]

<u>Answer:</u> The concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=3.82days$

Putting values in above equation, we get:

$k=\frac{0.693}{3.82}=0.181days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.181days^{-1}$

t = time taken for decay process = 3.00 days

$[A_o]$ = initial amount of the reactant = $1.45\times 10^{-6}mol/L$

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}$

$[A]=3.83\times 10^{-30}mol/L$

Hence, the concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

7 0

3 years ago

What is the molar solubility of agcl in 0. 30 m nacl at 25°C. ksp for agcl is 1. 77 × 10^-10.

Rzqust [24]

Molar solubility of AgCl will be 0.59 × $10^{-10}$ M.

The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.

A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.

Calculation of molar solubility:

Given data:

M = 0.30 M

$K_{sp}$ = 1.77 × $10^{-10}$

The reaction can be written as:

AgCl ⇔ $Ag^{+} + Cl^{-}$

s s (s+0.30)

$K_{sp}$ = [ $Ag^{+}$ ]+ [ $Cl^{-}$ ]

1.77 × $10^{-10}$ = s (0.30)

s = 1.77 × $10^{-10}$ / 0.3

s = 0.59 × $10^{-10}$ M

Therefore, molar solubility of AgCl will be 0.59 × $10^{-10}$ M.

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3 0

1 year ago

On molecular level what I causing the glaciers to melt

MakcuM [25]

Glaciers capture large amount of carbon dioxide from atmosphere. When concentration of carbon dioxide molecules in glaciers increase, then strength and fracture toughness of ice are decreased and <span>that make glaciers vulnerable to cracking and splitting into fragments.
</span>That is because hydrogen bonds between water molecules in glaciers is decreased under increasing concentrations of carbon dioxide who <span>competes with the water molecules connected in the ice crystal.</span>

6 0

3 years ago