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3 years ago

The Sun appears to be moving relative to what reference point?

Chemistry

1 answer:

gogolik [260]3 years ago

8 0

East to west. sometimes its northeast to northwest and sometimes is southeast to southwest.

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I need help with number 1 :/ please help me, thank you

Elena L [17]

Answer:

Amplitude is measured from the center line to the highest point in the waves.

Wavelength is the distance between one wave to the other from the highest point.

Frequency is the rate of the waves.

7 0

3 years ago

Why doesn't earth get hotter as it continues to receive radiation from the sun

Nataliya [291]

Because there is a balance between the amount of incoming radiation from the Sun and the amount of outgoing radiation from Earth.

hope this helps!

7 0

4 years ago

In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of wate

maxonik [38]

Answer:

21.344%

Explanation:

For the given chemical reaction, 8 moles of the reactant should produce 4 moles of $Na_{2}O$ . However, 195 g of $Na_{2}O$ was produced instead. The molar mass of $Na_{2}O$ is 61.9789 g/mol.

Thus, the moles of $Na_{2}O$ produced = 195/61.9789 = 3.1462 moles

The percent error = [(Actual -Experiment)/Actual]*100%

The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%

8 0

4 years ago

Which properties would be helpful in determining the identity of the liquids and which would not be helpful

rodikova [14]

Answer:

Extensive properties, such as mass and volume depend on the amount of substance present and are not useful in the identification of a substance. In this experiment, we will use three properties to identify a liquid substance: solubility, density and boiling point

3 0

1 year ago

A) A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M

Wewaii [24]

Answer:

i) 5.00 mL of 1.00 M NaOH: before the equivalence point

ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

iii) 100 mL of 1.00 M NaOH: at the equivalence point

iv) 150 mL of 1.00 M NaOH: after the equivalence point

v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

$M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol$

2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

$n=0.00500liter\times 1.00M=0.00500molNaOH$

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.

5 0

3 years ago