Atoms bond due to the interactions between

disa [49]

It depends on the plant you have. when you buy a new plant it usually has instructions so you would follow. Hope it helps!!!

3 0

3 years ago

Read 2 more answers

Why don't we use meters or miles when discussion distances between stars or galaxies?

taurus [48]

If we use mile or something else it will be hard to measure in sky .
Because the mile or meters too short for the space,
Galaxy and stars are too long far away from us if we use mile or meter,
It till be take like miliion year to measure.

4 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.}$

7 0

3 years ago

Measuring Volume

Dvinal [7]

Seems right, but if it’s just asking for one i would pick B

7 0

1 year ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t

Sedbober [7]

Answer:

Approximately $6.30\times 10^{-3}\;\rm mol$ .

Explanation:

The gallium here is likely to be produced from a $\rm NaGaO_2\, (aq)$ solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral " $\mathtt{(III)}$ " next to $\rm Ga$ . This numeral indicates that the oxidation state of the gallium in this solution is equal to $+3$ . In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of $\rm Ga\, (s)$ .

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium $\mathtt{(III)}$ solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

$t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s$ .

$Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C$ .

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

$\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}$ .

It takes three moles of electrons to deposit one mole of gallium atoms $\rm Ga\, (s)$ . As a result, $\rm 1.89\times 10^{-2}\; mol$ of electrons would deposit $\displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol$ of gallium atoms $\rm Ga\, (s)$ .

8 0

2 years ago