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3 years ago

Balanced equation for this? _ HBr + _ KHCO3  _ H2O + _ KBr + ___ CO2

Chemistry

1 answer:

Neporo4naja [7]3 years ago

8 0

HBr + KHCO3 ➡️ H2O + KBr + CO2
Everything is already balanced in the problem.

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Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.

goldfiish [28.3K]

Answer:Specific heat capacity at Constant volume of Nitrogen =742.32JKg.K

Explanation:

The Specific heat capacity at Constant volume of an ideal gas is given as

c =Cv / M

Cv= Constant volue of gas

M= Molar mass

But First, we determine the Constant volume, Cv which is given as

Cv = 5/2R since Nitrogen is a diatomic gas, N2 where R= 8.314Jmol.k

= 5/2 x 8.314Jmol.K

=20.785Jmol.K

Specific heat capacity at Constant volume, c

c= Cv/M

M=molar mass = 28.0 g/mol.

changing to kg/mol =28/1000= 0.028kg/mol

Therefore c =20.785Jmol.K/0.028kg/mol

=742.32JKg.K

8 0

3 years ago

A fuse is a safety device that stops the flow of electricity when a circuit overheats

allsm [11]

Answer:

Fuses are built as a safety measure if there is too much current flowing through a circuit. ... However, if an electrical current exceeds the maximum current allowed within that fuse it will overheat and cause the metal strip to melt. The destroyed metal strip breaks the connection and stops electricity flowing through.

7 0

3 years ago

Tissues are grouped together to form varius what?

Kruka [31]

the process goes like this.

1.cells

2.tissues

3.organs

4.organ systems

5.organisms

tissues are grouped to form organs.

I hope this helps.

4 0

3 years ago

A scientist has a new cough medicine by giving it to a group of course the scientist gives another group called the liquid and t

Liula [17]

Answer:50

Explanation:

Just look it up

6 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Balanced equation for this? ___ HBr + ___ KHCO3  ___ H2O + ___ KBr + ___ CO2

Balanced equation for this? _ HBr + _ KHCO3  _ H2O + _ KBr + ___ CO2