All categories

3 years ago

Question 4 (1 point) If you have exactly one million atoms how many moles do you have?

Chemistry

1 answer:

rjkz [21]3 years ago

7 0

Answer:

1.66 × 10⁻¹⁸ Moles

Explanation:

As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

Number of Moles = 1.66 × 10⁻¹⁸ Moles

You might be interested in

Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign,

stepan [7]

Answer:

0.37 %

Explanation:

Given that:

Calculated density of aluminum = 2.69 g/cm³

Accepted density of aluminum = 2.70 g/cm³

$Error\ percentage=\frac {|Accepted\ value-Calculated\ value|}{Accepted\ value}\times 100$

Thus, applying values as:

$Error\ percentage=\frac {|2.70-2.69|}{2.70}\times 100$

Percent error = 0.37 %

7 0

3 years ago

Read 2 more answers

The following are units for mass except _____. A. kg B. Mg C. mL D. cg

yarga [219]

Cg is not in units for mass

5 0

3 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

4 0

3 years ago

1.2 moles of (NH4)3PO3

Aleks04 [339]

1.2 moles of (nph4)3po3 is.......159.6 grams

3 0

2 years ago

Describe the day of June 21 at the North Pole in terms of daylight and darkness ( science )

frosja888 [35]

On June 21, as seen from the North pole ...

-- the sun has been up, and it's been light outside,
for the past three months ... ever since March 21 .

-- The sun won't set, and it won't be dark outside,
for another three months ... until September 21.

-- Here at the North pole, it stays daylight for six months straight.
Today, on June 21, we're exactly halfway through the period of
continuous daylight.

7 0

2 years ago