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3 years ago

How many grams are there in 7.4 x 1023 molecules of H2SO4?

Chemistry

1 answer:

bazaltina [42]3 years ago

4 0

7.4x10^23 = molecules of silver nitrate sample

6.022x10^23 number of molecules per mole (Avogadro's number)

Divide molecules of AgNO3 by # of molecules per mol

7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)

(I leave off the x10^23 because they both will divide out)

Use your periodic table to find the molar weight of silver nitrate.

107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3

Now multiply your moles of AgNO3 with your molar weight of AgNO3

1.229mol x 169.868g/mol = 208.767g AgNO3

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a sample of 25.0g of an unknown metal is added to 25.0ml of water in a graduated cylinder and the final volume is 28.5ml what is

leva [86]

Answer:

<h2>The answer is 7.14 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of metal = 25 g

volume = final volume of water - initial volume of water

volume = 28.5 - 25 = 3.5 mL

It's density is

$density = \frac{25}{3.5} \\ = 7.142857...$

We have the final answer as

Hope this helps you

8 0

2 years ago

300×175<br> Express your answer using the correct number of significant figures.

icang [17]

Answer:

300×175=52500

Explanation:

caculator

3 0

2 years ago

What mass of Cl2, in grams is contained in a 10.0L tank at 27oC and 3.50 atm pressure ? a. 1.42 grams b. 142 grams c. 1.01 kg d.

Evgen [1.6K]

Answer:

the correct answer is A

Explanation:

8 0

2 years ago

Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO

PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of $FCH_{2}COOH$

$FCH_{2}COOH(aq) + H_{2}O (l) FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)$

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = $[H_{3}O^{+}]$ = $[FCH_{2}COO^{-}]$ = 0.00770 M

$pK_{a} = 2.59K_{a} = 10^{-2.59} = 0.00257 M$

$K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}$

$0.00257 = \frac{0.00005929}{x - 0.00770}$

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0

3 years ago

1. How many atoms are there in 2.8 moles of carbon? You must show your work to receive credit.

Zolol [24]

Answer:

The answer to your question is

1.- 1.686 x 10²⁴ atoms

2.- 0.25 moles

Explanation:

1.-

1 mol ---------------- 6.023 x 10²³ atoms

2.8 moles ---------- x

x = (2.8 x 6.023 x 10²³) / 1

x = 1.686 x 10²⁴ atoms

2.- 1 mol ------------------ 6.023 x 10 ²³ molecules

x moles ------------- 1.50 x 10²³ molecules

x = (1.50 x 10²³ x 1) / 6.023 x 10²³

x = 0.25 moles

6 0

3 years ago

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