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Allisa [31]
3 years ago
5

What is the velocity of a fish travels 18 meters in 3 seconds

Chemistry
1 answer:
Leni [432]3 years ago
6 0
V=d/t
v=18m/3sec
v=6m/s
or 6 meters per second.
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Explain that : how we explore different between Licl and Nacl
Leona [35]

Answer:

See the explanation below, please.

Explanation:

In the bunsen burner, the gas and air inlet can be regulated manually. In the case of metals (such as lithium and sodium in this case) they contain an electron in the latter in its external electronic configuration. They are characterized by transferring electrons easily and produce the emission of light when excited, producing flames of different colors in the lighter (orange for sodium and red / scarlet for lithium)

8 0
3 years ago
One atomic model describes electrons moving in clouds around the nucleus.
bixtya [17]

Answer: D.  It is the currently accepted atomic model.

Explanation:

It is the mordern atomic model, also known as the Electron Cloud Model. Indicating that the nucleus of an atom is surrounded by a cloud of electrons.

8 0
3 years ago
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

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7 0
3 years ago
A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
aliina [53]

Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

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3 years ago
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