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cricket20 [7]
3 years ago
13

What is it that lives if it is fed, and dies if you give it a drink

Chemistry
2 answers:
nadezda [96]3 years ago
6 0

Answer:

ants if you splash a ant with sum water that shít is dead

Svetach [21]3 years ago
4 0

Answer:

fire

Explanation:

You might be interested in
Please help I will reward brainly
nordsb [41]
Your answer is yeast, since it is a single celled organism that can be classified into a subkingdom related to sac fungi.
6 0
4 years ago
CO_2 sublimes readily at 25°C. Which properties are usually associated with a compound that undergoes this kind of change?
antoniya [11.8K]

Explanation:

Sublimation is defined as a process in which solid state of a substance directly changes into vapor or gaseous state without undergoing liquid phase.

For example, naphthalene balls show sublimation at room temperature.

As this process does not cause any change in chemical composition of a substance. Hence, it is known as a physical process.

Similarly, when CO_{2} sublimes readily at 25^{o}C. This shows change in physical state of carbon dioxide is taking place, i.e, from solid to gaseous phase.

Thus, we can conclude that when CO_{2} sublimes readily at 25^{o}C then it means physical properties are usually associated with a compound that undergoes this kind of change.

4 0
3 years ago
What would be the mass of a 33.5dm3 sample of O2 at STP?
liraira [26]

Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

{ \tt{22.4 \:  {dm}^{3}   \: \dashrightarrow \: 16 \: grams}} \\  { \tt{33.5 \:  {dm}^{3}  \:  \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\  \\  \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}

7 0
3 years ago
PLEASE HELP!!!!!!!!!
andreev551 [17]

The average atomic mass of the imaginary element : 47.255 amu

<h3>Further explanation  </h3>

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

Mass atom X = mass isotope 1 . % + mass isotope 2.% ..

isotope E-47 47.011 amu, 87.34%

isotope E-48 48.008 amu, 6.895

isotope E-49 50.009 amu, 5.77%

The average atomic mass :

\tt avg~mass=0.8734\times 47.011+0.06895\times 48.008+0.0577\times 50.009\\\\avg~mass=41.059+3.310+2.886\\\\avg~mass=47.255~amu

5 0
2 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
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