= 3.14 × 10 with the power of 0
(scientific notation)

= 3.14e0
(scientific e notation)

= 3.14 × 100
(engineering notation)
(one)

= 3.14
(real number)

3 0

3 years ago

Reika [66]

Answer:

[ar]5s1401

Explanation:

nanbsbshdhshhshshshhd

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago

The specific heat of liquid water is 4.184 j/g· ?c. calculate the energy required to heat 10.0 g of water from 26.5?c to 83.7?c.

Roman55 [17]

Energy required=mass*specific heat*temperature change
=10*4.184*57.2
=2393.248j
=2.39*10^3

5 0

3 years ago