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Papessa [141]
3 years ago
12

MgCI2 how many element ?

Chemistry
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

2 elements magnesium(Mg) and chlorine (Cl)

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Ph is a scale that ____ the audacity or basicity of a solution along the range of zero to 14
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2 years ago
In an acid-base neutralization reaction 38.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution.
DaniilM [7]

Answer:

0.2 M.

Explanation:

  • For the acid-base neutralization, we have the role:

The no. of millimoles of acid is equal to that of the base at the neutralization.

<em>∴ (XMV) KOH = (XMV) H₂SO₄.</em>

X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),

M is the molarity.

V is the volume.

  • For KOH:

X = 1, M = 0.5 M, V = 38.74 mL.

  • For H₂SO₄:

X = 2, M = ??? M, V = 50.0 mL.

∴ M of H₂SO₄ = (XMV) KOH/(XV) H₂SO₄ = (1)(0.5 M)(38.74 mL)/(2)(50.0 mL) = 0.1937 M ≅ 0.2 M.

5 0
3 years ago
The half-life of gold-198 is 2.7 days. After
Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= \frac{x}{4} 

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{2.7days}=0.257days^{-1}

b) for completion of one fourth of reaction

t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}

t=\frac{2.303}{0.257}\log{4}

t=8.1days

Thus after 8.1 days , one fourth of original amount will remain.

8 0
3 years ago
What is the relationship between the distance between objects and the gravitational force between them
Leokris [45]

Answer: the gravitational pull decreases as the object's distance increases

Explanation:

3 0
3 years ago
Calculate the freezing point of a solution containing 5.7 % kcl by mass (in water).
laila [671]
5.7% KCl  is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g, 
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius

3 0
3 years ago
Read 2 more answers
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