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matrenka [14]
2 years ago
14

The point in a titration at which the indicator changes is called the

Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer:

End point

Explanation:

The point at which the indicator changes color is called the endpoint. So the addition of an indicator to the analyte solution helps us to visually spot the equivalence point in an acid-base titration

#correct me if I'm wrong

keep safe and study hard

brainliest please thank you

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What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
_H2 SiCl2 + _H2O −−→ _H8 Si4O4 + _HCl
Serjik [45]

Answer:

A

Explanation:

pls mark brainliest right

6 0
3 years ago
Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O
Bess [88]
40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
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5 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
3 years ago
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Option “A” is the crest of the wave because it’s the maximum value of upward displacement within a cycle.
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