HELP QUICK I HAVE A QUIZ ITS JUST MULTIPLE CHOICE

How many grams of dry nh4cl need to be added to 2.00 l of a 0.600 m solution of ammonia, nh3, to prepare a buffer solution that

Brrunno [24]

Nh4cl = 117.165 gram ( solution is attached )

8 0

3 years ago

The volume of a gas is 400.0 ml when the pressure is 1.00 atm. at the same temperature, what is the pressure at which the volume

almond37 [142]

C. 0.20atm (hope this is correct)

4 0

3 years ago

Read 2 more answers

The weight of the body decrease inside water why?

Lady_Fox [76]

Penurunan atau kehilangan massa otot bisa menimbulkan penurunan berat badan yang tidak direncanakan

3 0

3 years ago

"A crosslinked copolymer consists of 57 wt% ethylene (C2H4) repeat units and 43 wt% propylene (C3H6) repeat units. Determine the

ch4aika [34]

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = $\frac{57 g}{28 g/mol}=2.036 mol$

Moles of propylene = $\frac{43g}{42g/mol}=1.024 mol$

1 mole = $N_A =6.022\times 10^{23}$ molecules/ atoms

Units of ethylene = $2.036 mol\times N_A$

Units of propylene = $1.024 mol\times N_A$

a) Fraction of ethylene units:

$=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}$

b ) Fraction of propylene units:

$=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}$

3 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