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2 years ago

Wasting my points here who wants them?

Chemistry

2 answers:

Lubov Fominskaja [6]2 years ago

8 0

I would like them please I am new

yKpoI14uk [10]2 years ago

4 0

Me thanks for the points

You might be interested in

xenn [34]

The answer is OH.

Hope this helps!

4 0

3 years ago

What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound. That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms. The empirical formula is therefore N₂O.

8 0

3 years ago

To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i

enyata [817]

Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$ .....(1)

Molarity of acetic acid solution = 0.0880 M

Volume of solution = 30.6 mL

Putting values in equation 1, we get:

$0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol$

Thus there are 0.00269 moles of acetic acid in buffer.

8 0

3 years ago

9. Write a balanced nuclear equation for the following: The isotope Strontium-90 decays by Q-

Sergio039 [100]

Answer: $_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}$

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

General representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

General representation of alpha decay :

$_Z^A\textrm{Sr}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha$

The balanced nuclear equation when the isotope Strontium-90 decays by Q- decay is :

$_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}$

4 0

3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) = ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 % = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

6. Fórmula empírica: ZnO

3 0

3 years ago