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icang [17]
2 years ago
10

A sample of gold has a mass of 100 grams and a volume of 5cm^3, calculate the density by dividing the mass by volume

Chemistry
1 answer:
Dafna11 [192]2 years ago
7 0

Answer:

20cm^2

Explanation:

Here, Density= Mass/ Volume

=100/5

= 20 cm^2

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6 0
3 years ago
Which of the following is an incorrect name for an acid? A. acetic acid B. hydrocarbonate acid C. hydrocyanic acid D. sulfurous
just olya [345]

Answer:

The answer to your question is letter B

Explanation:

Incorrect name

A. acetic acid               This name is correct for the acid with formula CH₃COOH

B. hydrocarbonate acid  This is not the name for acid but for a molecule that has hydrogen and a metal.  

C. hydrocyanic acid   This name is correct for the inorganic molecule with formula HCN

D. sulfurous acid          This name is correct and is the name of the inorganic molecule with formula H₂SO₃.

E. phosphoric acid       This name is correct for the acid with formula H₃PO₄.

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3 years ago
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sasho [114]

Answer:

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6 0
2 years ago
If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0
2 years ago
Help needed Assignment is due Justify that H2SO4 is Arrhenius acid and KOH is Arrhenius base.
irakobra [83]

Answer:

<u><em>Arrhenius Acid:</em></u>

According to Arrhenius concept, Acids are proton donors.

Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>

<u><em>Arrhenius Base:</em></u>

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>

<u><em></em></u>

6 0
3 years ago
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