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Dimas [21]
3 years ago
13

3.95 g of sugar (C6H12O6) is dissolved in water to make 158 mL of solution. Find the molarity.

Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

[C₆H₁₂O₆] = 0.139 M

Explanation:

Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.

We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.

For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)

(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL

We determine moles, we convert them to mmoles, we divide by mL

M = 0.139 M

Moles = 3.95 g . 1mol / 180g → 0.0219 mol

We convert mL to L → 158 mL . 1L/1000mL = 0.158L

M = 0.0219 mol / 0.158L = 0.139 M

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How many grams of iodine are needed to prepare 28.6 grams of ICl
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The answer will be 17.9 grams
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Give explanations for the large drops in melting point from C to Si and from Ge to Sn.
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As we move down the group, the metallic bond becomes more stable and the formation of forming covalent bond decreases down the group due to the large size of elements.

Covalent and metallic bonding leads to higher melting points. Due to a decrease in attractive forces from carbon to lead there is a drop in melting point.

Carbon forms large covalent molecules than silicon and hence has a higher melting point than silicon.

Similarly, Ge also forms a large number of covalent bonds and has a smaller size as compared to that of Sn. Hence melting point decreases from Ge to Sn.

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3 0
1 year ago
A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
The density of solid ni is 8.90 g/cm3. how many atoms are present per cubic centimeter of ni
bulgar [2K]
Density gives mass of object per volume......   Here, density is given 8.90 g/cm3   therefore, per cubic centimeter contains 8.90 g Ni.   mole of Ni = mass / atomic mass   = 8.90 / 58.6934   = 0.1516 mole     number of atoms: mole * 6.022 * 10^23   = 0.1516 * 6.022 * 10^23   = 0.9129 * 10^23   = 0.9 * 10^23 (approx.)
7 0
3 years ago
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