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alexira [117]
3 years ago
11

hola todos que tengan un lindo día mi hijo quiere saber esta respuesta se lo daría yo pero tengo mucho trabajo y bueno se los de

jo a ustedes espero que se lo respondan bien, muchas gracias que tengan buenos días como dije antes saludos

Chemistry
1 answer:
Butoxors [25]3 years ago
6 0

Answer:

3.

(A) 1,000 gramos en un kilogramo.

(B) 8 kilogramos

(C) 1,500 gramos en un kilogramo y medio

(D) 5.5 kilogramos

4.

(A) 283 grados Kelvin

(B) 423 grados Kelvin

(C) 27 grados Celsius

(D) -3 grados Celsius

5.

Unidades de Tiempo

(A) 600 seg

(B) 420 min

(C) 12,600 seg

(D) 2,280 seg

Unidades de Longitud

(A) 2,000 m

(B) 12,000 m

(C) 7.587 km

(D) 5.5 m

Unidades de Masa

(A) 15,000 g

(B) 0.7 kg

(C) 12 kg

(D) 4,500 g

Unidades de Temperatura

(A) 302 K

(B) 121°C

(C) 343 K

(D) -8°C

6.

a) metros

b) centimetros

c) centimetros

d) kilometros

e) kilometros

7.

a) 14.76 m

b) 1,476 cm

8.

a) El auto mas rapido es el segundo. 45/5= 9 metros en un segundo.

b) El primero. El segundo recorre 7m en un segundo 35/5= 7.

9.

a) 100 centavos en 1 peso.

b) 1,060 centavos en 10.60 pesos.

c) 6.50 pesos

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A 5.00g quantity of a diprotic acid was dissolved in water and made up exactly 250 mL. Calculate the molar mass if the acid is 2
MAXImum [283]

Answer:

The molar mass of the diprotic acid is 90.10 g/mol.

Explanation:

The acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization.

H_2A+2KOH\rightarrow K_2A+2H_2O

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2  ( neutralization )

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of diprotic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=?\\V_1=25 mL\\n_2=1\\M_2=1.00 M\\V_2=11.1 mL

Putting values in above equation, we get:

2\times M_1\times 25=1\times 1.00\times 11.1}

M_1=\frac{1\times 1.00\times 11.1}{2\times 25}=0.222 M

Molarity of acid solution = 0.222 M =0.222 mol/L

Volume of original solution = 250 mL = 0.250 L ( 1 mL = 0.001 L)

Moles of diprotic acid in 0.250 L solution :

=0.222 mol/L\times 0.250 L=0.0555 mol

Mass of diprotic acid = m = 5.00 g

Moles(n)=\frac{mass(m)}{\text{Molar mass(M)}}

M=\frac{m}{n}=\frac{5.00 g}{0.0555 mol}=90.10 g/mol

The molar mass of the diprotic acid is 90.10 g/mol.

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