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2 years ago

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Help me real quick !!

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

yellow

if not sorry that's though the best answer

Explanation:

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Which of the following elements is a metalloid

klemol [59]

A metalloid can be:
- Boron (B)
- Silicon (Si)
- Germanium ( Ge)
- Arsenic (As)
- Antimony ( Sb)
- Tellurium (Te)
- Polonium (Po)

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6 0

3 years ago

Read 2 more answers

What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?

Zigmanuir [339]

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

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molar concentration

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3 years ago

What is the formula of the ionic compound formed by the elements lithium and oxygen?

kozerog [31]

The formula is Letter A) Li2<span>O

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6 0

2 years ago

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Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

2 years ago

Please help me with chemistry

Murljashka [212]

Answer:

60.052 g/mol (molar mass of vinegar)

84.007g/mol (baking soda )

7 0

2 years ago