Answer: Option b. 50 neutrons and 38 protons.
Explanation:
Atomic number = proton number = 38
Mass number = proton number + Neutron
Mass number = 88
Proton = 38
Neutron number =?
Neutron number = Mass number — proton number
Neutron number = 88 —38 = 50
In order from the most likely to bind an oxygen to least likely;
3 bound o2, po2=100mmhg1 bound o2, po2=100mmhg3 bound o2, po2=40mmhg<span>1 bound o2, po2=40mmhg
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Haemoglobin is more likely to bind oxygen if its other oxygen binding sites have already bound to an oxygen molecule. The higher the partial pressure of oxygen in the blood also makes it more likely that the hemoglobin will bind oxygen.
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Answer:
6.53g of K₂SO₄
Explanation:
Formula of the compound is K₂SO₄
Given parameters:
Volume of K₂SO₄ = 250mL = 250 x 10⁻³L
= 0.25L
Concentration of K₂SO₄ = 0.15M or 0. 15mol/L
Unknown:
Mass of K₂SO₄ =?
Methods:
We use the mole concept to solve this kind of problem.
>>First, we find the number of moles using the expression below:
Number of moles= concentration x volume
Solving for number of moles:
Number of moles = 0.25 x 01.5
= 0.0375mole
>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:
Mass(g) = number of moles x molar mass
Solving:
To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.
For:
K = 39g
S = 32g
O = 16g
Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)
= 78 +32 + 64
= 174g/mol
Using the expression:
Mass(g) = number of moles x molar mass
Mass of K₂SO₄ = 0.0375 x 174 = 6.53g
Answer:
1) Se2O5
2) I2O6
3)Zn3n2
4) Cr(HCO3)3
Explanation:
selenium pentaoxide (= also called diselenium pentoxide)
= Se2O5
⇒ Se = 78.97 g/mol
⇒ O = 16 g/mol
⇒ 2*78.97 + 5*16 = 237.94 g/mol
iodine trichloride
= I2O6
⇒ I = 126.9 g/mol
⇒ Cl = 35.45 g/mol
⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol
zinc (1) nitride does not exist (it's Zinc(ii)nitride
The oxidation number for zinc is always 2
Zn3n2
⇒ Zn = 65.38 g/mol
⇒ N = 14 g/mol
⇒3*65.38 + 2* 14 = 224.14 g/mol
chromium (III) bicarbonate
Cr(HCO3)3
⇒ Cr = 52 g/mol
⇒ H = 1.01 g/mol
⇒ C = 12 g/mol
⇒ O = 16 g/mol
52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol
.0071 you round 0 to1 because the 6 is bigger than 5 and if anything is bigger than five you round up that last number
