All categories

3 years ago

What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Chemistry

1 answer:

Neporo4naja [7]3 years ago

8 0

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

You might be interested in

What volume of hydrogen will be produced at STP by the reaction of 67.3 g of tin with excess water according to the following re

VikaD [51]

Answer:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

Explanation:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

5 0

3 years ago

Which statement summarizes the difference between mass and weight?

alexgriva [62]

Answer:

Mass is the amount of matter in an object.

Weight is how much an object weighs.

Hope this helps!

6 0

3 years ago

An unknown aqueous metal analysis yielded a detector response of 0.255. When 1.00 mL of a solution containing 100.0 ppm of the m

AVprozaik [17]

Answer:

1.022ppm is the unknown concentration of the metal

Explanation:

Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.

The unknown concentration (X) produces a signal of 0.255

99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502

0.99X + 1ppm produce 0.502, thus, X is:

0.255 * (0.99X + 1 / 0.502) =

X = 0.503X + 0.508

0.497X = 0.508

X =

1.022ppm is the unknown concentration of the metal

3 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago

g Alkenes and alkynes are called unsaturated compounds because ________. they have fewer hydrogen atoms attached to the carbon c

Firdavs [7]

Answer:

They have fewer hydrogen atoms attached to the carbon chain than alkanes

Explanation:

Let's compare ethane (an alkane) with ethene (an alkene) and ethyne (an alkyne):

Ethane's formula is C₂H₄, while ethene's is C₂H₄ and ethyne's C₂H₂.

As you can see, alkenes and alkynes have fewer hydrogen atoms attached to the carbon chain due to them having multiple bonds between the carbon atoms.

7 0

3 years ago