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3 years ago

If the caterpillar keeps moving at the same speed, how long will it take to travel 6cm?

Chemistry

1 answer:

Solnce55 [7]3 years ago

4 0

It will take 15 s to travel 6 cm

<h3>Further explanation</h3>

Given

distance versus time graph

Required

time travel

Solution

Caterpillar motion is a straight motion with a constant speed, so that the graph between distance and time forms a diagonal line

If we look at the graph, we can determine the time taken when the distance reaches 6 cm (y axis) by drawing a line to the diagonal line and cutting the x-axis as time, and we get 15 s

Or we can also use the formula for motion at constant speed:

d = v x t

With v at point 2,5 of 2/5 m / s, so the time taken:

$\tt t=\dfrac{d}{v}=\dfrac{6}{2/5}=15~s$

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In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(a

stiv31 [10]

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL$

$M_2=0.738M$

Thus, the concentration of the NaOH solution is, 0.738 M

7 0

3 years ago

About 6 × 109 g of gold is thought to be dissolved in the oceans of the world. If the total volume of the oceans is 1.5 × 1021 L

Lelechka [254]

First, determine the number of moles of gold.

Number of moles = $\frac{given mass in g}{molar mass}$

Given mass of gold = $6 \times 10^{9} g$

Molar mass of gold = 196.97 g/mol

Put the values,

Number of moles of gold = $\frac{6 \times 10^{9} g}{196.97 g/mol}$

= $0.03046\times 10^{9} mole$ or $3.046\times 10^{7} moles$

Now, molarity = $\frac{moles of solute}{volume of the solution in liters}$

Put the values, volume of ocean = $1.5 \times 10^{21} L$

Molarity = $\frac{3.046\times 10^{7} moles}{1.5 \times 10^{21} L}$

= $2.03\times 10^{-14} M$

Thus, average molar concentration = $2.03\times 10^{-14} M$

6 0

3 years ago

How many atoms are present in 34.96 moles of mg

vivado [14]

Multiply amount in grams by the relative atomic mass of magnesium then multiply by avagodro constant (6.02*10^23)

6 0

3 years ago

A closed vessel having a volume of 1 liter holds 2.5 × 1022 molecules of carbon dioxide gas. Determine the mass of CO2 present,

11111nata11111 [884]

Answer:

mass = 1.8x10⁻³ kg; number of moles = 4.1x10⁻⁵ kmol; specific volume = 0.55 m³/kg; molar specific volume = 24.4 m³/kmol

Explanation:

By the Avogadro's number, 1 mol of the matter has 6.02x10²³ molecules, thus, the number of moles (n) is the number of molecules presented divided by Avogadro's number:

n = 2.5x10²²/6.02x10²³

n = 0.041 mol

n = 4.1x10⁻⁵ kmol

The molar mass of CO₂ is 44 g/mol (12 g/mol of C + 2*16g/mol of O), and the mass is the number of moles multiplied by the molar mass:

m = 0.041 mol * 44 g/mol

m = 1.804 g

m = 1.8x10⁻³ kg

The specific volume (v) is the volume (1L = 0.001 m³) divided by the mass, and it represents how much volume is presented in each part of the mass:

v = 0.001/1.8x10⁻³

v = 0.55 m³/kg

The molar specific volume (nv) is the volume divided by the number of moles, and it represents how much volume is presented in each part of the mol:

nv = 0.001/4.1x10⁻⁵

nv = 24.4 m³/kmol

5 0

3 years ago

Hydrogen can be extracted from natural gas according to the following equilibrium.

antiseptic1488 [7]

Answer:

2.16x10⁻²

Explanation:

First, let's find out the molar concentrations of the reactants. The molar mass of CH4 is 16 g/mol, and of CO2 is 44 g/mol. The number of moles is the mass divided by the molar mass:

nCH4 = 24.0/16 = 1.5 moles

nCO2 = 88.0/44 = 2 moles

The concentration is the number of moles divded by the volume, thus:

[CH4] = 1.5/1 = 1.50 M

[CO2] = 2/1 = 2.00 M

For the equilibrium reaction, let's do an equilibrium chart:

CH4(g) + CO2(g) ⇄ 2CO(g) + 2H2(g)

1.50 2.00 0 0 Initial

-x -x +2x +2x Reacts (stoichiometry is 1:1:2:2)

1.50-x 2.00-x 2x 2x Equilibrium

As sateted in problem, [CH4] = 2.70*[CO]

1.50 - x = 2.70*2x

1.50 - x = 5.4x

6.4x = 1.50

x = 0.2344

Thus, at equilibrium:

[CH4] = 1.50 - 0.2344 = 1.2656 M

[CO2] = 2.00 - 0.2344 = 1.7656 M

[CO] = 2*0.2344 = 0.4688 M

[H2] = 2*0.2344 = 0.4688 M

The equilibrium constant is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients.

K = ([CO]²*[H2]²)/([CH4]*[CO2])

K = (0.4688²*0.4688²)/(1.2656*1.7656)

K = 0.0483/2.2345

K = 2.16x10⁻²

7 0

3 years ago