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4 years ago

5

On the period table, the rows are called

1 answer:

liubo4ka [24]4 years ago

5 0

They’re called periods, answer A

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Please help, i will mark you as brainliest

Mrrafil [7]

Answer:

The order of reactivity of metals is as follows, Potassium > Sodium > Lithium > Calcium > Magnesium > Aluminium > Zinc > Iron > Copper > Silver > Gold.

Explanation:

The reactivity of elements (metals) towards water decreases towards the right in a period. It also increases down the group. But zinc is more reactive towards water than iron. Hence the correct order is:

Iron<Zinc<Magnesium<Sodium

4 0

2 years ago

Read 2 more answers

calculate the mass of 120cc nitrogen present at STP. how many number of molecules are present in it?

Stells [14]

Answer:

$0.15008\ \text{g}$

$3.23\times 10^{21}$

Explanation:

1 mol of nitrogen at STP = 22.4 L = 22400 cc

n = Mol of $N_2$ = $\dfrac{120}{22400}=0.00536\ \text{mol}$

M = Molar mass of $N_2$ = $28\ \text{g/mol}$

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

Mass of $N_2$ is

$m=nM\\\Rightarrow m=0.00536\times 28\\\Rightarrow m=0.15008\ \text{g}$

Mass of the nitrogen is $0.15008\ \text{g}$

Number of molecules is given by

$nN_A=0.00536\times 6.022\times 10^{23}=3.23\times 10^{21}\ \text{molecules}$

The number of molecules present in it are $3.23\times 10^{21}$

5 0

3 years ago

How do you use the changes in the phase of water to keep you cool?

sineoko [7]

Answer:

by freezing

Explanation:

8 0

3 years ago

What animal wins march mammal madness in 2021?

Semmy [17]

Answer:

turtles i like turtles

Explanation:

6 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago