arrange the following in order of increasing mass . a. 16 water molecules b. 2 atoms of lead 3. 5.1*10^-23

(3.88 x 10to the 26th)atoms of copper would be equal to how many moles of copper?

marissa [1.9K]

Answer:

69

Explanation:

hey u said what I said didusbevehdysg

4 0

2 years ago

Who discovered significance of 6.022 x 10 to the 23 power

Masja [62]

6.02 x 10^23 is Alvogrado's number (Amedio Alvogrado is the name). This formula is determined to find the amount of particles within a substance by a mole.
I hope this helps!

6 0

3 years ago

A researcher dilutes 30.0 ml of ethanol to 300.0 ml with distilled water. what is the percentage concentration by volume of the

Len [333]

Answer: 9.09 %

Explanation:

To calculate the percentage concentration by volume, we use the formula:

$\text{Volume percent of solution}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100$

Volume of ethanol (solute) = 30 ml

Volume of water (solvent) = 300 ml

Volume of solution= volume of solute + volume of solution = 30+ 300 = 330 ml

Putting values in above equation, we get:

$\text{Volume percent of solution}=\frac{30}{30+300}\times 100=9.09\%$

Hence, the volume percent of solution will be 9.09 %.

3 0

3 years ago

Read 2 more answers

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0

3 years ago

Which pair of elements have the same valence electronic configuration of np³?

Dennis_Churaev [7]

Answer:

(c) P and Sb

Explanation:

We can determine the number of valence electrons of an element:

If it belongs to Groups 1 and 2, the number of valence electrons is equal to the number of group and the differential electron occupies the s subshell.

If it belongs to the groups 13-18, the number of valence electrons is equal to: "Number of group - 10" and the differential electron occupies the p subshell.

Which pair of elements have the same valence electronic configuration of np³?

(a) O and Se. NO. They belong to the group 16 and the valence electron configuration is ns² np⁴.

(b) Ge and Pb. NO. They belong to the group 14 and the valence electron configuration is ns² np².

(c) P and Sb. YES. They belong to the group 15 and the valence electron configuration is ns² np³.

(d) K and Mg. NO. They belong to the groups 1 and 2 and the valence electron configuration is ns¹ and ns².

(e) Al and Ga. NO. They belong to the group 13 and the valence electron configuration is ns² np¹.

5 0

3 years ago

arrange the following in order of increasing mass . a. 16 water molecules b. 2 atoms of lead 3. 5.1*10^-23​

arrange the following in order of increasing mass . a. 16 water molecules b. 2 atoms of lead 3. 5.1*10^-23