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stiv31 [10]
3 years ago
15

The density of a sample of gold is 19.32 g/cm'. If the sample has a volume of 48.9 cm', what is the mass?

Chemistry
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

\boxed {\tt 944.748 \ grams}

Explanation:

The formula for density is:

d=\frac{m}{v}

Rearrange the formula for mass, m. Multiply both sides by v.

d*v=\frac{m}{v} *v

d*v=m

Mass is found by multiplying the mass by the volume.

The density of the gold is 19.32 grams per cubic centimeter. The volume is 48.9 cubic centimeters.

d= 19.32 \ g/cm^3 \\v=48.9 \ cm^3

Substitute the values into the formula.

m= 19.32 \ g/cm^3 *48.9 \ cm^3

Multiply. Note the centimeters cubed will cancel out.

m= 19.32 \ g * 48.9

m= 944.748 \ g

The mass of the gold is 944.748 grams.

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In each of the following sets of elements, which one will be least likely to gain or lose electrons?
klasskru [66]
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).

2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.

3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.

4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
7 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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gavmur [86]

Answer:

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Explanation:

5 0
3 years ago
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chubhunter [2.5K]

Answer:

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Explanation:

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nataly862011 [7]

Answer:

Option B, Because of the reversible nature of crystallizing and dissolving

Explanation:

Solution containing the maximum amount of solute that can be dissolved in the given solvent at the particular temperature is called saturated solution.

Reversible reaction is the reaction which can go in reverse and forward direction both on varying reaction condition.

In the saturated NaCl solution, on lowering temperature, The the dissolved NaCl molecules may crystallize. Likewise on increasing temperature, the crystallized crystals may dissolved. As the reaction moves in both the direction, therefore its considered to be equilibrium system.

Therefore, amog given, option B is correct.

Because of the reversible nature of crystallizing and dissolving

8 0
3 years ago
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