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3 years ago

fish give off the compound ammonia which has a pH above 7 to which class of compounds does ammonia belong

Chemistry

1 answer:

jasenka [17]3 years ago

5 0

Answer:
The compound ammonia given by fish is alkaline

Explanation:
We can classify elements/compounds based on their pH values into three types:
acids: these are compounds having pH value lower than 7
neutral: these are compounds having pH value equal to 7
alkalies: these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)

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i don't need these solved I just need the steps to solve it and. what these problems in chemistry would be called so I can look

wolverine [178]

The answers are as follows,

1. 4.71 L,

2. 3.29 L,

3. 1634.6 torr

Explanation:

All the above problems can be sorted out using the Boyle's law represents the relationship between the volume and pressure.

It can be expressed as follows,

P1V1 = P2V2

Rearranging the above expression, we get,

$$ V2 = \frac{P1V1}{V2}$

1) The first case can be solved as,

$$V 2=\frac{921 \mathrm{mm} \mathrm{Hg} \times 3.89 \mathrm{L}}{760 \mathrm{mm} \mathrm{Hg}}=4.71 \mathrm{L}$

2) The second case can be solved as,

$$V 2=\frac{25 L \times 600 \text { torr}}{4560 \text { torr}}=3.29 L$

3) The third case can be solved as,

$$P 2=\frac{7.43 L \times 1100 \text { torr}}{5 L}=1634.6 \text { torr}$

Thus the answers for all the three cases are found as,

1. 4.71 L,

2. 3.29 L,

3. 1634.6 torr respectively

6 0

3 years ago

Please help me with this. And show all work as well ASAP!!

qaws [65]

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

$p_1=x_1p_{total}$

where, x = mole fraction

$p_{total}$ = total pressure

$x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}}$ ,

$x_{oxygen}=\frac{3.0}{12.33}=0.243$ ,

$p_{oxygen}=0.243\times 770torr=187torr$

Thus the partial pressure of oxygen is 187 torr.

6 0

3 years ago

Under which conditions of temperature and pressure would a 1-liter sample of a real gas behave most like an ideal gas?

Sladkaya [172]

Answer : The correct option is (3) 500 K and 0.1 atm.

Explanation :

A real gas behaves ideally at high temperature and low pressure.

The ideal gas equation is,

$PV=nRT$

where,

P = pressure of gas

V = Volume of gas

R = Gas constant

T = temperature of gas

n = number of moles of gas

The ideal gas works properly when the inter-molecular interactions between the gas molecules and volume of gas molecule will be negligible. This is possible when pressure is low and temperature is high.

Therefore, the correct option is (3) 500 K and 0.1 atm.

3 0

4 years ago

Read 2 more answers

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago

Determining Density and Using Density to Determine Volume or Mass

Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = $\frac{mass}{volume}$

ρ=m/v ,m=ρv, v=m/ρ

(a)m=1*10g , v=7.36cm³

ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

v= 65/0.791 =82.17g/mL

m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

ρ=374.5/41.8 =8.96g/cm³

mass of ethanol=15g, density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0

4 years ago