How many grams of solute are present in 1250 mL of a 1.34 M NaNO3 solution?

Se tiene un litronde solucion de agua con alcohol con una consentracion del 4.5% v/v ¿Cuanto alcohol se tiene en la solución

Vinvika [58]

Answer:

45 mL

Explanation:

Tenemos los siguientes datos:

V = 1 L

C = 4,5% v/v

El porcentaje en volumen (%v/v) expresa el volumen de soluto (alcohol en este caso) que hay cada 100 mL de solución. Si la solución tiene una concentración del 4,5% v/v eso quiere decir que hay 4,5 ml de alcohol cada 100 ml de solución, de acuerdo a lo siguiente:

4,5% v/v alcohol = volumen alcohol/ volumen solución x 100 = 4,5 mL alcohol/100 mL solución= 4,5 mL alcohol/0,1 L alcohol

Por lo tanto, al multiplicar por el volumen total de la solución (1 L), obtenemos la cantidad total de alcohol:

4,5 mL alcohol/0,1 L alcohol x 1 L = 45 mL

3 0

3 years ago

6)

scZoUnD [109]

Answer:

A.) Linear

Explanation:

4 0

3 years ago

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

Calculate the mass, in grams, of cucl2 (mw = 134.452 g/mol) required to prepare 250.0 ml of a 6.11 % w/v cu2 (mw = 63.546 g/mol)

Sever21 [200]

6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution

therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g

# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles

1 mole of CuCl2 contain 1 mole of Cu2+ ion

Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2

Molar mass of CuCl2 = 134.452 g/mole

The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams

6 0

3 years ago

A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

kupik [55]

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

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6 0

3 years ago