What is the name of The compound?

What is the boiling point for water in degrees celsius?

soldier1979 [14.2K]

Answer:

<h2>100°C is the boiling point of water in degrees Celsius</h2>

4 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

How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?

Kitty [74]

Answer:

21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C7H14 + 21O2 —> 14CO2 + 14H2O

From the balanced equation above, 2L of C7H14 produced 14L of H2O.

Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.

Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2O will occupy

26.25L i.e

Xmol of H2O = 26.25/22.4

Xmol of H2O = 1.172 mole

Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

Number of mole H2O = 1.172 mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O =..?

Mass = mole x molar mass

Mass of H2O = 1.172 x 18

Mass of H2O = 21.10g

Therefore, 21.10g of H2O is produced from the reaction.

3 0

3 years ago

Bae wya .......... please bae I can't find you please show up

Elan Coil [88]

Answer:

Your babe is in your school

Explanation:

go ask babes out

7 0

2 years ago

Read 2 more answers

What is the molar mass of glucose, C3H603?

ziro4ka [17]

Answer:

The mass percentage of carbon can be found easily using the molar mass of C6H12O6, 180.1559 g/mol. We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass. C6H12O6 = CO2 + C3H6O3 + CH3OCH3 Take fructose for example. Compound.

Explanation: I looked it up

4 0

2 years ago