Answer:
MgSO4.7H2O
Explanation:
Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O
Mass of the hydrated salt (MgSO4.xH2O) = 12.845g
Mass of anhydrous salt (MgSO4) = 6.273g
Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g
Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:
Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x
Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt
18x/120 + 18x = 6.572/12.845
Cross multiply to express in linear form
18x x 12.845 = 6.572(120 + 18x)
231.21x = 788.64 + 118.296x
Collect like terms
231.21x — 118.296x = 788.64
112.914x = 788.64
Divide both side by 112.914
x = 788.64 /112.914
x = 7
Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O
A pH of 2 indicates a acid
Answer:
THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.
AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.
Explanation:
The question above follows Boyle's law of the gas law as the temperature is kept constant.
Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Mathematically, P1 V1 = P2 V2
P1 = 150 kPa = 150 *10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 kPa * 2.48 / 2.98
P2 = 372 *10 ^3 / 2.98
P2 = 124.8 kPa.
The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.
The answer is GRAINS
i realy hope this helped
