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laila [671]
3 years ago
13

Name two elements that have the same properties as calcium (Ca). (4 points)

Chemistry
1 answer:
enyata [817]3 years ago
7 0
 beryllium, magnesium, strontium,barium<span>, and </span><span>radium. </span>
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How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?
stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using  ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

7 0
3 years ago
Which molecule contains the smallest number of hydrogen atoms? A) N2H4. B) Al(OH) 3. C) (NH4) 2SO4. D) NH4NO3.
Vesna [10]
The molecule that contains the fewest number of Hydrogen atoms would be B. Al(OH)3. It only has 3 Hydrogen atoms.
7 0
2 years ago
Read 2 more answers
Which choice below would match homozygous recessive?
LUCKY_DIMON [66]
The answer is B

Homozygous mean the same and you know that the two alleles will be the same (either BB or bb) and receive is usually the lower case set of alleles
8 0
2 years ago
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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
calculate the mass required to prepare 2.5 L of 1.0 M NaOH solution. Given that the molar mass for NaOH is 40 g/mol.
Helen [10]

Answer:

The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g

Explanation:

We do this by preparing the equation:

Mass = concentration (mol/L) x volume (L) x Molar mass

Mass = 1.0 M x 2.5 L x 40 g/mol

Mass = 100 g

3 0
2 years ago
Read 2 more answers
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