All categories

3 years ago

Select whether the statement is for Speed, Velocity, or Acceleration.

Chemistry

1 answer:

Vesna [10]3 years ago

4 0

Answer:

I think that the statement is relative to speed because it is saying km per second.

You might be interested in

Water flows over Niagara Falls at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing

Zinaida [17]

Answer:

1) The power of Niagara Falls is 1.176 × 10⁹ W

2) The number of 15 W LED light bulbs it could power is 78.4 × 10⁶ light bulbs

Explanation:

1) The Niagara falls water mass flow rate = 2,400,000 kg/s

The height of the fall = 50 meters

The gravitational potential energy = Mass (kg) × height (m) × gravity (9.8 m/s²)

The power = The energy converted per second = Mass flow rate (kg/s) × height (m) × gravity (9.8 m/s²)

Therefore;

The power of Niagara Falls= 2,400,000 kg/s × 50 m ×9.8 m/s²= 1.176 × 10⁹ W

The power of Niagara Falls = 1.176 × 10⁹ W

2) The number, n, of 15 W LED light bulbs it could power is given by the relation;

n × 15 W = 1.176 × 10⁹ W

∴ n = 1.176 × 10⁹ W/(15 W) = 78.4 × 10⁶ light bulbs

The number of 15 W LED light bulbs it could power = 78.4 × 10⁶ light bulbs.

6 0

3 years ago

How many grams of the parent isotope are left in the sample after three half lives?

pochemuha

After 3 half-lives, 125 grams of the parent isotope will remain.

5 0

3 years ago

Describe a way to separate iron filings from sand.<br><br> Enter your answer in the space provided.

Lina20 [59]

Use the magnetic force of iron filings

5 0

3 years ago

Read 2 more answers

What are watts and how do they work

ipn [44]

A watt is a unit of power.

4 0

3 years ago

Read 2 more answers

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

4 0

3 years ago