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3 years ago

Which metal element is most reactive Beryllium, magnesium, calcium, or strontium

Chemistry

2 answers:

Rudiy273 years ago

4 0

I think the answer is magnesium. I hope this helps!

777dan777 [17]3 years ago

3 0

Answer:

Magnesium I’m pretty sure

Explanation:

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List the following types of electromagnetic radiation in order of increasing wavelength.

chubhunter [2.5K]

Answer:

Radio, Microwave, Infared, Ultraviolet

Explanation:

look at the EM spectrum

3 0

3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

Make a plot of arsenate species vs. pH for a 25 mM arsenate solution. Vary pH from 0-14. Neglect activity corrections. A groundw

dusya [7]

!!!!!!!!!!!!!!!!!!!!

5 0

3 years ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

i: van 't Hoff factor (1 for non-electrolytes)
Kf: cryoscopic constant
m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0

2 years ago

Help Please! Thanks!

soldier1979 [14.2K]

Answer:

Explanation:

it would be the most reasonable

4 0

2 years ago

Read 2 more answers