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2 years ago

Guys someone shared their notes with me how should I thank them

Chemistry

2 answers:

Rzqust [24]2 years ago

8 0

by answering thiers questions also then be kind to share

Dvinal [7]2 years ago

6 0

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Complete the statement to describe the calcium concentration that must be maintained in human body cells. ( fill in the blank )T

erik [133]

Answer:

The calcium concentration must be greater outside the cell than inside the cell.

Explanation:

My previous answer was deleted from the explanation I provided from another website.

7 0

4 years ago

Read 2 more answers

You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

<u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

8 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

Based on the diagram below, how much of the excess reactant is left over? *

Alexxandr [17]

Answer:

3 of lunchmeat and 2 slices of cheese

Explanation:

From the question given,

Each sandwich contains:

2 bread + 3 lunch meat + 1 cheese.

Now, the limiting reactant is the slice of bread.

We can determine the leftover as follow:

1. For the lunchmeat.

From the simple equation above,

2 bread requires 3 lunchmeat.

Therefore, 6 bread will require = (6 x 3)/2 = 9 lunchmeat.

Lunchmeat given = 12

Lunchmeat required = 9

Leftover lunchmeat = 12 – 9 = 3

Therefore, 3 lunchmeat is leftover.

2. For cheese.

From the simple equation above,

2 bread requires 1 cheese.

Therefore, 6 bread will require = 6/2 = 3 cheese.

Cheese given = 5

Cheese required = 3

Leftover cheese = 5 – 3 = 2

Therefore, 2 cheese is leftover.

From the simple illustrations above,

3 lunchmeat and 2 cheese are leftover.

7 0

3 years ago

Magnesium oxide (MgO) forms when the metal burns in air. (a) If 1-25 9 of MgO contains 0.754 g of Mg, what is the mass ratio of

Vsevolod [243]

Answer:

Explanation:

a )

1.25 g MgO contains .754 g of Mg .Rest will be O

so oxide = 1.25 - .754 = 0.496 g

ratio of magnesium to oxide = .754/.496 = 1.52

b) 1.25 g of MgO contains .754 g of Mg

534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g

5 0

3 years ago