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2 years ago

If a dog has a mass of 16.1 kg, what is its mass in the following units? Use scientific notation in all of your answers.

Chemistry

1 answer:

nekit [7.7K]2 years ago

3 0

Answer:

$16.1\times 10^3\ \text{g}$

$16.1\times 10^{6}\ \text{mg}$

$16.1\times 10^{9}\ \mu\text{g}$

Explanation:

Mass of dog = $16.1\ \text{kg}$

$1\ \text{kg}=10^{3}\ \text{g}$

$1\ \text{kg}=10^{6}\ \text{mg}$

$1\ \text{kg}=10^{9}\ \mu\text{g}$

So,

$16.1\ \text{kg}=16.1\times 10^3\ \text{g}$

$16.1\ \text{kg}=16.1\times 10^{6}\ \text{mg}$

$16.1\ \text{kg}=16.1\times 10^{9}\ \mu\text{g}$

Mass of the dog in the other units is $16.1\times 10^3\ \text{g}$ , $16.1\times 10^{6}\ \text{mg}$ and $16.1\times 10^{9}\ \mu\text{g}$ .

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Under a certain set of conditions, the percent yield of a reaction that produces carbon dioxide is 75.0%. What mass in grams of

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Answer:

Actual yield = 20.03g

Explanation:

$% yield = 100.\frac{actual yield}{theoretical yield}$ % $yield$ = $100. \frac{actual \ yield}{theoretical \ yield}$

Actual yield = % yield.(theoretical yield) / 100

Actual yield = 75.26.7/100

Actual yield = 20.025

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2 years ago

Which statement is true at STP? (The atomic mass of Zn is 65.39 u.)

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3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

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8 0

3 years ago

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

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3 years ago