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Firdavs [7]
3 years ago
12

If m = 45g and V = 15ml, what is D in g/ml?​

Chemistry
1 answer:
ch4aika [34]3 years ago
6 0

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

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According to VSEPR theory , in which fashion will the bonds and lone pairs of electrons be arranged about the central atom in th
Elina [12.6K]

Answer:

Tetrahedral, trigonal pyramidal, trigonal bipyramidal.

Explanation:

The VSPER theory states that the bonds of sharing electrons and the lone pairs of electrons will repulse as much as possible. So, by the repulsion, the molecule will have some shape.

In the ion PO₄³⁻, the central atom P has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons at the valence shell and needs 2 to be stable. 3 oxygens share 1 pair of electrons with P, and the two lone pair remaining in P is shared with the other O, then the central atom makes 4 bonds and has no lone pairs, the shape is tetrahedral.

In the ion H₃O⁺, the central atom O has 6 electrons in its valence shell and needs 2 electrons to be stable. The hydrogen has 1 electron, and need 1 more to be stable. The hydrogens share 1 pair of electrons with the oxygen, then it remains 3 electrons at the central atom, and the VSPER theory states that the shape will be a trigonal pyramidal.

In the AsF₅, the central atom As has 5 valence electrons, and F has 1 electron in its valence shell, so each F shares one pair of electrons with As, and there are no lone pairs in the central atom. For 5 bonds without lone pairs, the shape is trigonal bipyramidal.

7 0
3 years ago
Starch and cellulose are both produced by plants, yet one is easily digested by animals and the other is not. Discuss the differ
Nadusha1986 [10]
<span>Starch and cellulose have the same substance but different structures. They are both polysaccharides. The basic unit of a polysaccharide is the glucose. Glucose, which contains carbon, hydrogen, and oxygen, have two forms. The alpha-glucose with an alcohol group attached to carbon 1 is down and the beta-glucose with the alcohol group attached to carbon 1 is up. Starch is the alpha-glucose while cellulose is the beta-glucose. Starches are linked into a straight chain whereas the cellulose are connected like a pile of stack paper. When the human body eats starch, it can digest the starch but not the cellulose because it has no enzyme that can break it down. </span>
4 0
3 years ago
Doing Science plz help
Murljashka [212]

Answer:

My guess is b or c but its robably wrong

Explanation:

I just also need points sorry <3

4 0
3 years ago
Read 2 more answers
The law of constant composition states: All atoms of a given element have a constant composition and are different than atoms of
Airida [17]

Answer:

C

Explanation:

The law proves C. For examples no matter how water you have it will always have a 1:2 ratio of oxygen to hydrogen. :)

7 0
2 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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