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3 years ago

How many moles are there in 5.30 x 1024 atoms of silver?

Chemistry

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

The mass of

4.6

atoms of silver is approximately 820 g.

Explanation:

In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.

mole atoms Ag=6.022xx10

atoms Ag

Molar mass of Ag =#"107.87 g/mol"#

Multiply the given atoms of silver by

mol Ag

6.022

atoms Ag

. Then multiply times the molar mass of silver.

4.6

atoms Ag

mol Ag

6.022

atoms Ag

107.87

g Ag

mol Ag

820 g Ag

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What are sister chromatids

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3 years ago

Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

Which of the following is a precipitation reaction?

fredd [130]

Answer:

B is a precipitation reaction.

Explanation:

This is because a precipitation reaction is when a solid is made from the combination of cations and anions in a solution to create a solid.

3 0

3 years ago

Read 2 more answers

How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?

Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

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3 years ago

Read 2 more answers

What concentration of monosodium phosphate would be required to make a solution of pH 7.4 with 0.2 M disodium phosphate ( pKa

denis-greek [22]

Answer:

The concentration of monosodium phosphate is 0.1262M

Explanation:

The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2

To determine the pH you must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].

For H₂PO₄⁻ / HPO₄⁻ buffer:

pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]

As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:

7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]

0.2 = log [0.2] / [H₂PO₄⁻]

1.58489 = [0.2] / [H₂PO₄⁻]

[H₂PO₄⁻] = 0.1262M

<h3>The concentration of monosodium phosphate is 0.1262M</h3>

8 0

3 years ago