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3 years ago

A) 30.22 g NaCl x 1 mol NaC l = 58.4430 Molar mass (g) NaCl=0.5171 mol NaCl

1 answer:

6 0

The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.

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What is the mass of the block of iron illustrated below?

Mandarinka [93]

VOLUME= 5cm*10cm*2cm =100cm^3
but density of iron=7.874g/cm^3
mass=7.874g*100 =787.4g
mass of that block = 787.4g

7 0

3 years ago

Read 2 more answers

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

Anna007 [38]

Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8

1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))

COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm

5 0

3 years ago

A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency

saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f

f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0

2 years ago

3 points

NNADVOKAT [17]

Answer:

Explanation:

cause

4 0

3 years ago

Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.

Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

4 0

3 years ago