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3 years ago

6

Create a time that marks important events during the Precambrian Supereon.

2 answers:

Maurinko [17]3 years ago

8 0

I really hope this help!!

sesenic [268]3 years ago

4 0

Multiple Answers:

The earth started cooling and the outer edge of the planet solidified from molten lava to a solid crust. Water rained from the atmosphere and created oceans. The first form of life on our planet was created during the Archean eon in these oceans.

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Help Plzz Will give all points

ale4655 [162]

Answer:

reqirured mediem

Explanation:

5 0

2 years ago

Read 2 more answers

Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the sol

Afina-wow [57]

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

$Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}$

So,

$Number\ of\ moles = Molarity\ \times Volume\ in\ L$

$= 0.223\times 0.141$

= 0.031 moles

Now

$Mass = moles \times Molecualr\ weight$

where,

The Molecular weight of NaCl is 58.44 g/mole

And, the moles are 0.031 moles

So, the mass of NaCL is

$= 0.031 \times 58.44$

= 1.811 g

We simply applied the above formulas

3 0

2 years ago

Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati

Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S$ = change in entropy

$\Delta H_{vap}$ = change in enthalpy of vaporization = 40.5 kJ/mol

$T_b$ = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

$\Delta S=\frac{40.5kJ/mol}{352K}$

$\Delta S=115J/mol.K$

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0

3 years ago

Convert 1000mg=__g 1L=__mL 160cm=__mm 1.4km=__m 109g=__kg 250m=__km 80cm=__m 75mL=__L 5.6m=__cm 6.5g=__mg 170.4m=__cm 564Dg=__g

Snowcat [4.5K]

Answer:

1000mg= 1g

1L= 1000 mL

160cm = 1600mm

1.4km= 1400m

109 g = 0.109kg

250m= 0.250 km

80cm= 0.8 m

75mL= 0.075L

5.6m= 560 cm

6.5g= 6500mg

170.4m= 17040 cm

564 Dg = 5640 g

58 dg = 5800 mg

600 L= 0.6 KL

0.0923Km= 92300 mm

Explanation:

1 mg = 1x10⁻³ g

1 g = 1000 mg

1 g = 10 dg

1 g = 1x10⁻³ kg

1 Dg = 10 g

1 dg = 100 mg

1 L = 1000 mL

1 L = 1x10⁻³ KL

1 mL = 1x10⁻³ L

1 km = 1000 m

1 km = 1x10⁶ mm

1 m = 1x10⁻³ km

1 cm = 1x10⁻² m

1 cm = 10 mm

5 0

3 years ago

How can u separate saturated, unsaturated and supersaturated solution?

Kruka [31]

Answer:

Saturated = The solution cannot dissolve any more solute at a given temperature

2) Unsaturated = solution can dissolve more solute at a given temperature.

3) Supersaturated = Solution which has more solute than its saturated solution

Explanation:

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4 0

2 years ago