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elena-14-01-66 [18.8K]

3 years ago

11

How many oxygen atoms are there in the following compound: 2HNO3 3 5 6 12

2 answers:

Fudgin [204]3 years ago

7 0

Answer:

6

Explanation:

There is 2 of the coumpond, in one compound, there are 3 oxygens. But when there are two, it is 6.

Ahat [919]3 years ago

3 0

Answer:

The answer is there are 3 oxygen atoms.

Explanation:

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Which of these elements is unlikely to form covalent bonds? s, h, mg, ar, si?

kari74 [83]

<span>Answer: K because it is metal and typically forms ionic bonds. Ar is also unlikely to form any bonds because it has a full outer shell of electrons, but it can form covalent bonds.</span>

4 0

3 years ago

Read 2 more answers

CaO + H2O -> Ca(OH)2

yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075$

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

$\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g$

% yield :

$\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%$

8 0

2 years ago

What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use

ch4aika [34]

Explanation:

The given data is as follows.

$M_{1}$ = 10 mM = $10 \times 10^{-3}$ M

$V_{1}$ = 750 ml, $V_{2}$ = 5 ml

$M_{2}$ = ?

Therefore, calculate the molarity of given NaCl stock as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml$

$M_{2}$ = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.

7 0

3 years ago

?Al + ?H2SO4 → ?Al2(SO4)3 +3H2

AfilCa [17]

The answer is: 27 grams of aluminium.

Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

n(H₂) = 1.5 mol; amount of hydrogen.

Form chemical reaction: n(Al) : n(H₂) = 2 : 3.

n(Al) = 2 · 1.5 mol ÷ 3.

n(Al) = 1.0 mol; amount of aluminium.

m(Al) = n(Al) · M(Al).

m(Al) = 1 mol · 27 g/mol.

m(Al) = 27 g; mass of aluminium.

5 0

3 years ago

A 4.00g sample of helium has a volume of 24.4L at a temperature of 25.0oC and a pressure of 1.00 atm. The volume of the helium i

Masteriza [31]

Answer:

0.41 moles.

Explanation:

Given that:

Mass of helium = 4.00 g

Initial Volume = 24.4 L

initial Temperature = 25.0 °C =( 25 + 273) = 298 K

initial Pressure = 1.00 atm

The volume was reduced to :

i.e

final volume of the helium - 10.4 L

Change in ΔV = 24.4 - 10.4 = 10.0 L

Temperature and pressure remains constant.

The new quantity of gas can be calculated by using the ideal gas equation.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{1.00*10.0}{0.082057*298}$

n = 0.4089 moles

n = 0.41 moles.

7 0

3 years ago