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just olya [345]
3 years ago
13

Examples of noble gases

Chemistry
1 answer:
LekaFEV [45]3 years ago
4 0
Natural gas, desolate
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Two groups were tasked to measure the mass of the metal objects given by their teacher. Rachel’s group has the following data fo
Studentka2010 [4]

The measurement of Rachel’s group is precise but not accurate while the measurement of Ashley’s group is accurate but not precise.

Precision has to do with how close together the values obtained from a scientific measurement is. If we take a look at the values obtained by  Rachel’s group, we will notice that the values are exactly 1.00 g apart. This means that the values are precise.

However, these values a far from the actual value which is 100.00 g therefore the measurement of Rachel’s group is precise but not accurate.

On the other hand, the values obtained by Ashley’s group are; 99.5 g, 100.1 g, and 100.5 g. These values are very close to the actual value which is 100.00 g  hence they are accurate.

The values obtained by Ashley’s group do not have consistent intervals therefore, they are not precise.

Learn more; brainly.com/question/15664210

5 0
3 years ago
How many moles of lithium metal would need to react with excess oxygen gas to produce 4 moles of lithium oxide in the following
irina1246 [14]

8 moles of Lithium are needed.

3 0
3 years ago
Which temperature most likely corresponds to the asthenosphere?
iren2701 [21]

Answer:

A. 1613 °C

Explanation:

The asthenosphere is located just below the Earth’s crust.

It must be molten or semi-molten, so that convection currents can move the tectonic plates above it.

Rocks start to melt at about 1300 °C, so a temperature of 1613 °C sounds about right.

4 0
3 years ago
Read 2 more answers
Choices for images
BabaBlast [244]

One

Let's start by stating what we know is wrong. Equilibrium is achieved when the reactants and products have a stable concentration. That makes D incorrect. Equilibrium is not established until about the 6th or 7th second.

The fact that you get any products at all means that the reactants will become products.  Just who is favored has to be looked at very carefully. The products start very near 0. They go up until their concentration at equilibrium. When the reach equilibrium, the products have increased to 17. The reactants have dropped from 40 to 27. By a narrow margin, I would say the products are favored.

C is incorrect. There are still reactants left.  

E is incorrect. the reactants started out with a concentration of 40. The reaction is not instantaneous.  The concentration was highest at 40 or right at the beginning. This assumes that the reactants were mixed and the products were produced and the water/liquid amount has not changed.

B is incorrect. The concentration of the reactants is higher at equilibrium.

A is wrong. It is product favored.

I'm getting none of the above.

Problem Two

AgBr is insoluble (very). You'd have to work very hard to get them to separate into their elemental form. Just putting AgBr in water isn't enough. Lots of heat and lots of electricity are needed to get the elemental form.

I suppose you should pick B. Mass must be preserved. But if you balanced the equation, it would work with heat and electricity.



3 0
3 years ago
Read 2 more answers
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ \frac{5.60}{2}gallons = 2.80 gallons

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = (M__1}) = Density*Volume

                          = 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}

                         = 11659.06grams

On the other hand; the mass of water = (M__2})= Density*Volume

                         = 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}

                        = 10577.95 grams

Molarity = \frac{massof solute*1000}{molarmassof solute*massofwater}

              =  \frac{11659.06*1000}{62.07*10577.95}

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

\Delta T_{boiling} = k_{boiling}M

where,

k_{boiling} = 0.512 °C/m

\Delta T_{boiling} =  100°C + 17.56 × 0.512

              = 109.09 °C

6 0
3 years ago
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