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Dennis_Churaev [7]

3 years ago

6

For the chemical equations shown below, label each reactant as either acid or base, and each product as either conjugate acid or

conjugate base according to the brønsted-lowry definition. drag each label to the appropriate target.

1 answer:

Veronika [31]3 years ago

8 0

Acids are ionic compounds with hydrogen at the front. Bases are ionic compounds with hydroxide at three back(bronsted definition)
A conjugate acid is a base that has gained a proton(hydrogen ion) and a conjugate base is formed when an acid has given up its proton.

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If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t

Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is 49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

<u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g$

<u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g$

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 120 g

$m_2$ = mass of cold water = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of hot water = 95°C

$T_2$ = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=49^oC$

Hence, the final temperature of the mixture is 49°C

4 0

3 years ago

Which of these atoms are electrically neutral?

Xelga [282]

Atoms are electrically neutral because they contain equal quantities of positively charged protons and negatively charged electrons. Electrons and protons have equal but opposite charges, so the result is no net charge. Ions are atoms that have gained or lost electrons.

4 0

3 years ago

Read 2 more answers

The chart shows the voltage of four electric currents.

masha68 [24]

Answer: The correct option is Current W flows at a higher rate than Current Z.

Explanation: To answer this question, we will require Ohm's law.

Ohm's Law states that the current flowing through a conductor across two points is directly proportional to the voltage difference across that two points.

Mathematically,

$V=IR$

where, V = voltage

I = Current

R = resistance

For the given question, assuming that the resistance is constant. So, the current is directly proportional to the voltage.

$V\propto I$

Hence, as the current W is greater of all the given currents so, it will flow at a higher rate.

Therefore, the correct answer is Current W flows at a higher rate than Current Z.

7 0

3 years ago

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calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

<u>Answer:</u>

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0

2 years ago

What do you call a solid which dissolves?

san4es73 [151]

Answer:

solute

Explanation:

for example sugar salt etc

solute dissolve in solvent

8 0

2 years ago