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Answer:
151.63 g
Explanation:
We first get the number of moles;
Moles = Molarity × volume
= 2.0 × 0.475
= 0.95 moles
1 mole of CuSO4 = 159.609 g/mol
Therefore;
Mass = moles × molar mas
= 0.95 moles × 159.609 g/mol
= 151.63 g
Your answer should be Polysaccharides
MARK ME BRAINLIEST PLEASE!!!!!!!
Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3
question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest
Answer:
78.2 g/mol
Step-by-step explanation:
We can use the <em>Ideal Gas Law</em> to solve this problem:
pV = nRT
Since n = m/M, the equation becomes
pV = (m/M)RT Multiply each side by M
pVM = mRT Divide each side by pV
M = (mRT)/(pV)
Data:
ρ = 2.50 g/L
R = 0.082 16 L·atm·K⁻¹mol⁻¹
T =98 °C
p = 740 mmHg
Calculation:
(a)<em> Convert temperature to kelvins
</em>
T = (98 + 273.15) = 371.15 K
(b) <em>Convert pressure to atmospheres
</em>
p = 740 × 1/760 =0.9737 atm
(c) <em>Calculate the molar mass
</em>
Assume V = 1 L.
Then m = 2.50 g
M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)
= 76.14/0.9737
= 78.2 g/mol
