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3 years ago

6

Sea floor spreading occurs at a mid-ocean ridge true or false

1 answer:

Elenna [48]3 years ago

7 0

Answer:

False

Explanation:

sea floor spreading is not consistent at all mid ocean ridges.

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Which unit conversion allows you to convert 21.0 gallons into milliliters?

Ulleksa [173]

Answer:

Explanation:

we are given

there are in 0.073 gallons

we can see that there are two non-zero places after decimals

so, there are two siginificant figures

step-1: Change gallons to quarts

we know that

1 gallon = 4 quarts

so, we can write as

step-2: Change into liter

we know that

1 liter = 1.057 quarts

or

step-3: Change into millilitre

we know that

Since, it is asking for only two significant figures

so, we can also write as

...................Answer

3 0

3 years ago

The box in the picture begins motionless, and then both forces are applied at the same time. Describe the resulting motion of th

vaieri [72.5K]

It seems that you have missed to attach the given image for us to answer this question, but anyway, here is the answer. Based on the picture of a box which is motionless, the resulting motion of the box would be option B. The box will move to the left. Hope this answer helps.

3 0

3 years ago

Read 2 more answers

Which has the highest boiling point .33 m NH3 or .10 m Na2SO4

Aleksandr-060686 [28]

Answer:

0.33 mol/kg NH₃

Explanation:

Data:

b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.

1 mol NH₃ ⟶ 1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

5 0

3 years ago

A compound is made up of iron and oxygen, only. The ratio of iron ions to oxide ions is 2:3 in this compound. The IUPAC name for

never [62]

Answer: The correct option is 3.

Explanation: We are given a compound which is made up of iron and oxygen only. The ratio of the two are given as:

$\frac{\text{iron ions}}{\text{oxide ions}}=\frac{2}{3}$

This means that, number of iron ions are 2

Number of oxide ions are 3

From the above information, the formula becomes : $Fe_2O_3$

The valency of iron = 3

Valency of oxide = 2

This compound is named as iron (III) oxide.

Hence, the correct option is 3

6 0

3 years ago

Read 2 more answers

A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v

miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution $C_1=?$

Volume of the sport drink before dilution $V_1=5.4 mL$

Concentration of red dye in sport drink after dilution $C_2=18.1 ppm$

Volume of the sport drink after dilution $V_2=25.0 mL$

$C_1V_1=C_2V_2$ ( dilution )

$C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}$

$C_1=83.80 ppm\approx 84.0ppm$

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0

3 years ago