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3 years ago

Balance the following equation _MgF2 + _Li2CO3 -> _MgCO3 + _LiF

Chemistry

1 answer:

malfutka [58]3 years ago

3 0

Answer:

1 MgF2 + 1 Li2CO3 --> 1 MgCO3 + 2 LiF

Explanation:

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Answer:

CH4 + 2O2 ----> CO2 + 2H2O

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3 years ago

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Copper(II) oxide reacts with carbon when heated.

Alexus [3.1K]

Answer:

the fourth one or the first, but I wouldn't say I'm completely sure.

Explanation:

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3 years ago

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a faucet is leaking water at a speed of 5 drops per minute. If 1 ml is equivalent to 10 drops, how many liters of water will be

nikdorinn [45]

Answer: 0.72 litres of water is wasted in one day.

Explanation:

First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72

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3 years ago

Measurement of heat conductivity A metal panel of area A = 100 cm² and thickness Y = 0.5 cm was found to conduct heat at a rate

RSB [31]

Answer:

The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.

Explanation:

The heat flow through this area of the metal panel can be expressed as:

$q/A=-k\frac{(T_2-T_1)}{Y}$

being A the surface area, k the thermal conductivity, T2 and T1 the temperatures in each side and Y the thickness of the metal panel.

We have to rearrange the equation to clear k:

$k=\frac{-(q/A)*Y}{\Delta T} = \frac{-(3 W/100 cm2)*0.5cm}{5 ^\circ K}*(\frac{1m}{100cm} )^{3} \\\\k=\frac{0.015}{5*100^{3} }W/mK= 3*10^{-9}W/mK$

The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.

7 0

3 years ago

You titrated a 25.00 mL solution of 0.02 M oxalic acid with a freshly prepared solution of KMnO4. If it took 42.32 mL of this so

RideAnS [48]

Answer:

This is a Redox Titration reaction and the overall ionic equation is given as follows;

2 MnO⁴⁻ + 5C₂O₄ ²⁻+ 16H+ → 2Mn²⁺ + 10CO₂ + 8H₂O

From here, we could see that the stoichiometry ratio of KMnO₄ and oxalic acid = 2:5

To calculate the molarity of KMnO₄, we will use the following equation.

x₁M₁V₁= x₂M₂V₂ (1)

Where,

x₁ = 2, stoichiometry number of KMnO₄ in the balanced reaction

x₂ = 5, stoichiometry number of C₂O₄ in the balanced reaction

M₁ and M₂ are the molarities of oxalic acid and KMnO₄ solutions

V1 and V2 are the volumes of oxalic acid and KMnO₄ solutions.

Making M₂, the subject of the formula and substituting the given values, we have,

2M₁V₁ = 5M₂V₂, M₂=2M₁V₁ / 5V₂

= = 2x0.02Mx25mL/ 5x42.32mL

M₂=0.0047M

Hence, the molarity of the KMnO4 at endpoint =0.0047M

5 0

4 years ago