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a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine

iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction $\frac{a}{b}$

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

$\frac{15.39 g of gold}{2.77 g of chlorine}$

The proportion is the equal relationship that exists between two reasons and is represented by: $\frac{a}{b}=\frac{c}{d}$

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

$\frac{15.39 g of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}$

Solving for the mass of gold gives:

$mass of gold=1 g of chlorine*\frac{15.39 g of gold}{2.77 g of chlorine}$

mass of gold= 5.56 grams

So, there is 5.56 g of gold for every 1 g of chlorine

5 0

3 years ago

Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and Tr

ale4655 [162]

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

$=\frac{x^2}{0.02 -x}$

$=8.32 \times 10^{-9}$

Clearing $x$ , we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$ , one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

$= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.

3 0

3 years ago

Please help

aev [14]

Answer:

Copper(II) sulphate – sodium hydroxide reaction

The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.

5 0

3 years ago

What particle limitation to the widespread of this reusable water bottles? Pls help someone

Sauron [17]

Answer: Plastic water bottles

Explanation:

If you use disposable water bottles, here are some important concerns you should know about how they’re made as well as the problems they cause for the planet, your health, and your wallet.

3 0

3 years ago

The pHof a buffer solution containing 0.10 Macetic acid and 0.10 M

zysi [14]

Chemical reaction: CH₃COO⁻(aq) + H⁺(aq) ⇄ CH₃COOH(aq).

H⁺ is from HNO₃: HNO₃ → H⁺ + NO₃⁻.

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.

8 0

3 years ago