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dangina [55]
2 years ago
10

If the following ions Mg2+, Cl-, Sr2+, OH- are placed in a test tube, the precipitate formed is

Chemistry
1 answer:
Solnce55 [7]2 years ago
3 0

Answer:

I am so sorry if I am wrong but my guess is no precipitate is formed.

Explanation:

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What prediction can i make about malaria
Sunny_sXe [5.5K]

Answer:

A disease caused by a plasmodium parasite, transmitted by the bite of infected mosquitoes.

The severity of malaria varies based on the species of plasmodium.

Symptoms are chills, fever, and sweating, usually occurring a few weeks after being bitten.

People traveling to areas where malaria is common typically take protective drugs before, during, and after their trip. Treatment includes antimalarial drugs.

Explanation:

4 0
3 years ago
How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?
lara [203]
<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

  • <em>q</em> is heat (in J)
  • <em>m</em> is mass (in g)
  • <em>c</em> is specific heat (in J/g °C)
  • ΔT is change in temperature (in °C or K)

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
  2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
  3. Multiply:                                                                                                           q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0
3 years ago
Use the periodic table to identify each of the following bonds as polar or nonpolar.
Molodets [167]

Answer:

Polar  

Step-by-step explanation:

Electronegativity increases from <em>left to right</em> in the Periodic Table.

Cl is further right than C (both tinted pink) in the portion of the Periodic Table below.

Cl is <em>more electronegative</em> than C, so the Cl has a partial negative charge and the C has a partial positive charge.

The C-Cl bond is polar.

5 0
3 years ago
Which event best helped Becquerel determine uranium radiates rays?
Elenna [48]
(D) exposure of a photographic plate
5 0
3 years ago
Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )
Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

4 0
3 years ago
Read 2 more answers
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