If Sally has 250 grams of cement, how many kilograms does she have? How many milligrams?

Why the mass of a rusted nail is greater then the nail before it rusted?

kifflom [539]

This is the reaction formula,
4Fe+3O2=2Fe2O3
3Fe+202=Fe3O4
it has oxygen atom after it's rusted

6 0

3 years ago

Some plants can reproduce both sexually and asexually. Explain one benefit of this dual reproduction capability.

pantera1 [17]

Answer:

An additional advantage of asexual reproduction is that colonization of new habitats may be easier when an individual does not need to find a mate to reproduce. During sexual reproduction the genetic material of two individuals is combined to produce genetically diverse offspring that differ from their parents.

Explanation:

An organism has many benefits for having the ability to reproduce both sexually and asexually. The first of these benefits is that it doesn't have to find a mate in order to reproduce, so it can create offspring by itself, and enable the continuation of its' species.

If you need more help on this subject, don't be afraid to ask me. I'm willing to help.

8 0

2 years ago

Molar mass of A1C1 3

Kamila [148]

Answer:

Gold(III) chloride

5 0

2 years ago

At what temperature (°C) will a 10.00 g sample of neon gas exert a pressure of 96.7 kPa in a

Eva8 [605]

Answer : The temperature of neon gas will be, 221.0 K

Explanation :

To calculate the temperature of neon gas we are using ideal gas equation.

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of neon gas = 96.7 kPa = 0.955 atm

Conversion used : (1 atm = 101.3 kPa)

V = volume of neon gas = 9.50 L

T = temperature of neon gas = ?

R = gas constant = 0.0821 L.atm/mole.K

w = mass of neon gas = 10.00 g

M = molar mass of neon gas = 20 g/mole

Now put all the given values in the ideal gas equation, we get:

$(0.955atm)\times (9.50L)=\frac{10.00g}{20g/mole}\times (0.0821L.atm/mole.K)\times T$

$T=221.0K$

Therefore, the temperature of neon gas will be, 221.0 K

7 0

2 years ago

How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add

Nostrana [21]

Answer :

The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g

Given : pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

Step 1 : To find pka

H₂PO₄⁻ <=> HPO₄²⁻

The above reaction has pka = 7.2 ( from image shown )

Step 2 : Plug values in Hasselbalch- Henderson equation .

Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :

$pH = pka + log\frac{[A^-]}{[HA]}$

pH = 6.86 pKa = 7.2

$6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Subtracting both side by 7.2

$6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Removing log

$10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$ ---------------- equation (1)

Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M

Assume [H₂PO₄⁻ ] = x

So , [x ] + [ HPO₄²⁻ ] = 0.1 M

[ HPO₄²⁻ ] = 0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]

[H₂PO₄⁻ ] = x

[ HPO₄²⁻ ] = 0.1 - x

Equation (1) = > $\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$

Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :

$\frac{[0.1 - x ]}{[ x]} = 0.457$

Cross multiplying

$0.1 - x = 0.457 x$

Adding x on both side

$0.1 -x + x = 0.457 x + x$

$0.1 = 1.457 x$

Dividing both side by 1.457

$\frac{0.1}{1.457} = \frac{1.457 x }{1.457}$

x = 0.0686 M

Hence , [H₂PO₄⁻ ] = x = 0.0686 M

[ HPO₄²⁻ ] = 0.1 - x

[ HPO₄²⁻ ] = 0.1 - 0.0686

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L

Since that volume of buffer solution is 1 L , so Molarity = mole

Hence Mole of NaH₂PO₄ = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄

Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :

Molar mass of Na₂HPO₄ = $141.96 \frac{g}{mol}$

Molar mass of NaH₂PO₄ = $119.98 \frac{g}{mol}$

$Mass (g) = mole (mol)* molar mass(\frac{g}{mol})$

$Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}$

Mass of Na₂HPO₄ = 4.457 g

$Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}$

Mass of NaH₂PO₄ = 8.23 g

5 0

2 years ago