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ANEK [815]
3 years ago
10

Both equations I and II are balanced, but equation I is the correct way to write the balanced equation.

Chemistry
1 answer:
ElenaW [278]3 years ago
5 0
I don’t get the question. is there a worksheet?
You might be interested in
How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

3 0
3 years ago
Object A has a mass of 3,600 kilograms. Object B has a mass of 600 kilograms. The weight of Object B will be ________ times the
Natalija [7]

Hi the answer is 1/6, because 1/6 of 3,600 is 600, so it's 1/6. Have a great day!


7 0
3 years ago
Elaborate on the nuclear model of the atom.
GarryVolchara [31]

Answer:

D

Explanation:

The nuclear model of the atom was proposed by Ernest Rutherford.

In his submission, the atom consists of a dense positive nucleus with electrons moving round this nucleus as planets move round the sun. He likened his model to the solar system. The nucleus is at the center of this system just as the sun is at the center of the solar system.

4 0
3 years ago
What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×
Ira Lisetskai [31]
^{10}_5 B
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}

The answer is A. 1.67 × 10⁻²⁶ kg.
8 0
3 years ago
calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate
kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

#SPJ4

4 0
1 year ago
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