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3 years ago

How many grams of NaCl (molecular weight = 58 Da) would you have to dissolve in 50ml to make a 2M stock solution? Answer:

Chemistry

1 answer:

Lerok [7]3 years ago

7 0

Answer:

5.8 g

Explanation:

Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.

The amount of NaCl required is calculated as follows:

(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol

This amount is converted to grams using the molar mass (58 g/mol).

(0.1 mol)(58 g/mol) = 5.8 g

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A solution was made by dissolving 5.10 mg of hemoglobin in water to give a final volume of 1.00 ml. the osmotic pressure of this

Assoli18 [71]

The molecular weight of hemoglobin can be calculated using osmotic pressure

Osmotic pressure is a colligative property and it depends on molarity as

πV = nRT

where

π = osmotic pressure

V = volume = 1mL = 0.001 L

n = moles

R = gas constant = 0.0821 L atm / mol K

T = temperature = 25°C = 25 + 273 K = 298 K

Putting values we will get value of moles

$moles=\frac{\pi V}{RT}=\frac{0.00195X0.001}{0.0821X298}mol$

we know that

$moles=\frac{mass}{molarmass}$

Therefore

$molarmass=\frac{mass}{moles}=\frac{5.10X10^{-3}g}{7.97X10^{-8}}=6.399X10^{4}g$

5 0

3 years ago

Which of the following is NOT an important physical property of matter?

IgorLugansk [536]

The best option is melting point

4 0

3 years ago

How many millimeters are in 251 cm?

antoniya [11.8K]

Answer:

2510

Explanation:

Since there are 10 milimetres in 1 centimetre, 1 cm = 10mm

There are 10 times as many milimetres as centimetres.

251cm X 10 = mm

251cm = 2510mm

3 0

3 years ago

Read 2 more answers

What mass of Na2SO4 is needed to make 2.5L of 2.0M solution? (Na=23g; S=32g; O=16g)

zaharov [31]

The molar mass of Na₂SO₄ -
2 x Na - 2 x23 = 46
1 x S - 1 x 32 = 32
4 x O - 4 x 16 = 64
total = 46 + 32 + 64 = 142 g/mol
the molarity of solution - 2.0 M
in 1 L of solution , 2.0 moles
Therefore in 2.5 L - 2 mol/L x 2.5 L = 5 mol
then the mass of Na₂SO₄ required = 142 g/mol x 5 mol = 710 g

8 0

3 years ago

Read 2 more answers

PLEASE HELP AND EXPLAIN!Calculate the energy required to heat 187.0 g of ammonia from 36.1 °C to 52.9 °C. Assume the specific he

Alenkinab [10]

Answer:

$E=1.48x10^4J$

Explanation:

Hello!

In this case, since the energy implied in a heating process is computed by using the following equation:

$E=mC(T_f-T_i)$

Whereas m is the mass, C the specific heat and T the temperature. In such a way, by plugging in the given mass, specific heat and temperatures, we obtain the following energy:

$E=187.0g*4.703\frac{J}{g\°C} (52.9\°C-36.1\°C)\\\\E=1.48x10^4J$

Considering that the specific heat can by used by unit of °C or K because their difference is equivalent.

Regards!

3 0

2 years ago