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3 years ago

Determine the molecular weight for Mg(NO3)2

1 answer:

4 0

The molecular weight is the same as molecular mass. You find this by using the periodic table
The molar mass of Mg(NO3)2 is 148.3

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What is the percent composition by mass of chlorine in LiCI

Lelechka [254]

Answer: Lithium (Li) 16.373%

Chlorine (Cl) 83.627%

Explanation:

6 0

2 years ago

Which predator is typically found in Central Africa (but occasionally at geocode 34.043021, -118.2668356) and slays its opponent

noname [10]

I believe it is a black mamba.

6 0

3 years ago

You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1

jeka57 [31]

Answer:

$\boxed{\text{36 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O$

2. Calculate the moles of HCOOH

$\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}$

3. Calculate the moles of NaOH.

$\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}$

4. Calculate the volume of NaOH

$c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}$

3 0

3 years ago

Given the following balanced equation:

svetlana [45]

So you would just take the coefficient and put it into ratio form

Answers:
1) 4:5
2)4:6
3)6:4
4)6:6
5)6:4

3 0

2 years ago

If 1.9 kJ of heat is transferred to 96 g aluminum at 113°C, what would the

erastova [34]

Answer:

T2 = 135.1°C

Explanation:

Given data:

Mass of water = 96 g

Initial temperature = 113°C

Final temperature = ?

Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)

Specific heat capacity of aluminium = 0.897 j/g.°C

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

ΔT = T2 - T1

Now we will put the values in formula.

Q = m.c. ΔT

1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C

1900 j = 86.112 j/°C × T2 - 113°C

1900 j / 86.112 j/°C = T2 - 113°C

22.1°C + 113°C = T2

T2 = 135.1°C

4 0

3 years ago