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3 years ago

How many atoms are in 0.230grams Pb

Chemistry

1 answer:

Brilliant_brown [7]3 years ago

5 0

Its about 11.5hg because if you divide it with the atom it would result to 11.5hg

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When 25.0 grams of solid potassium hydroxide (KOH, molar mass = 56.1 g/mol) is dissolved in 100.0 grams of water, the solution i

Maru [420]

Answer:

They gave you the equation; Cp=,

just plug everything in! You’ve seen this; I have long ago, but we had different units. Sorry, but it’s right there! Go get it!

Explanation:

4 0

3 years ago

There are molecules of carbon dioxide (CO2) in 102.5 grams.

Gekata [30.6K]

Answer:

1.403x10²⁴ molecules

Explanation:

In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first convert grams to moles using its molar mass:

102.5 g ÷ 44 g/mol = 2.330 mol CO₂

Now we convert moles into molecules using Avogadro's number:

2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules

7 0

3 years ago

An ideal gas sample is confined to 3.0 L and kept at 27 °C. If the temperature is raised to 77 °C and the initial pressure was 1

Nadusha1986 [10]

The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .

<h2>Hope it helps you please mark as brainlist</h2>

6 0

2 years ago

How many valence electrons does magnesium (Mg) have?Question options:A) 6B) 2C) 12D) 0

wel

INFORMATION:

We must find the number of valence electrons for magnesium

STEP BY STEP EXPLANATION:

In order to know the number of valence electrons for Mg, we need to locate the element in the periodic table

Since Mg is in the second group of the periodic table, it has two valence electrons.

ANSWER:

B) 2

4 0

1 year ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago