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SVETLANKA909090 [29]
3 years ago
12

A 5.40 mL sample of an H3PO4 solution of unknown concentration is titrated with a 5.000×10−2 M NaOH solution. A volume of 7.02 m

L of the NaOH solution was required to reach the equivalence point.
What is the concentration of the unknown H3PO4 solution?
Express your answer using four significant figures.
Chemistry
1 answer:
uysha [10]3 years ago
8 0

Answer:

0.02167 M

Explanation:

At first we need balanced equation

3NaOH + H3PO4 = Na3PO4 + 3H2O

This means for every 3 moles of NaOH we have 1mol of H3PO4.

At equivalence point we have equal moles of both titrant and analyte in our case H3PO4 and NaOH. start by finding the moles of NaOH

Molarity= moles/Litres

0.05= y/0.00702

x= 3.51x10^-4 moles

But keep in mind the balanced equation that says for every 3 moles of NaOH we have 1mol of H3PO4

so we divide by 3 moles to get moles of h3po4.

3.51x10^-4 moles NaOH x 1mol of H3Po4/3moles of NaOH

= 1.17x10^-4 moles of H3PO4

to find the molarity then use the forumla above

M = moles/litres

= 1.17x10^-4/0.0054

= 0.02167 M

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Sphinxa [80]

Answer:

b) F

Explanation:

Only non-metals can gain electrons during ionic bonding.

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4 0
4 years ago
How many moles at STP would take up 44.8 liters?
Tom [10]

Answer:

n =  2 mol

Explanation:

Given data:

Pressure = standard = 1 atm

Temperature = standard = 273.15 K

Volume = 44.8 L

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Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 44.8 L = n × 0.0821 atm.L/ mol.K  × 273.15 K

44.8 atm.L = n × 22.43 atm.L/ mol

n = 44.8 atm.L /   22.43 atm.L/ mol

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4 0
3 years ago
in order to make tea, 32,000 J of energy were added to 100.0g of water. what was the temperature Chang of the water? ​
mina [271]

Answer:

ΔT  = 76.5 °C

Explanation:

Given data:

Amount of water = 100.0 g

Energy needed = 32000 J

Change in temperature = ?

Solution,

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = m.c. ΔT

ΔT  = Q / m.c

ΔT  = 32000 j/ 100.0 g × 4.184 j/g. °C

ΔT  =  32000 j / 418.4 j /°C

ΔT  = 76.5 °C

5 0
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Sergio [31]

Answer:

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Explanation:

Hello, in this case, one could consider the undergoing chemical reaction as:

SrCO_3(s)\rightleftharpoons SrO(s)+CO_2(g)

Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:

Kp=p_{CO_2}^{eq}=1.59atm

Hence, since at the beginning there is no carbon dioxide, its pressure at equilibrium equals Kp:

Kp=p_{CO_2}^{eq}=1.59atm

Which was clearly defined above.

Best regards.

8 0
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irinina [24]

Answer:

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