Answer:
Yes they woll react.
Explanation:
Ferous oxide and zinc will be product.
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.
To know more about hydrate compound, visit the below link:
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Using the chart that has been provided, we may determine water temperature. We do this by drawing a straight line form the bottom scale which has the ppm of oxygen dissolved to the middle scale which has the percentage saturation.
The line starts from 11.5 ppm on the bottom scale and goes to 90% on the middle scale. Next, we continue this line, without changing its slope, to the third scale showing temperature. We see that it crosses the temperature scale at 4°C.
The temperature of the water is 4 °C.
Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Answer:
[H3O+] = 1.0*10^-12 M
[OH-] = 0.01 M
Explanation:
We can use the following equation to find the hydronium ion concentration. Plug in the pH and solve for H3O+.
pH = -log[H3O+]
<u>[H3O+] = 1.0*10^-12 M</u>
Now, to find the hydroxide ion concentration we will use the two following equations.
14 = pH + pOH
pOH = -log[OH-]
14 = 12 + pOH
pOH = 2
2 = -log[OH-]
<u>[OH-] = 0.01 M</u>
