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3 years ago

15 POINTS!!!! Please answer 1-8

Chemistry

1 answer:

Effectus [21]3 years ago

4 0

1. combustion
2. succession
3. condensation
4. precipitation
5. pioneer species
6. decomposition
7. evaporation
8. nitrogen fixation

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21<2x+3 help fast please

kolezko [41]

Answer:

Explanation:

x in (-oo:+oo)

2 < (1/2)*x-3 // - (1/2)*x-3

2-((1/2)*x)+3 < 0

(-1/2)*x+2+3 < 0

5-1/2*x < 0 // - 5

-1/2*x < -5 // : -1/2

x > -5/(-1/2)

x > 10

x in (10:+oo)

(10:+oo)

7 0

3 years ago

Read 2 more answers

The density of a metal is 9.80 g/mL. What is the mass of a sample of metal when dropped in 28.9 mL of water, the volume increase

yan [13]

The correct answer is 2.70 × 10² g or 270 g.

It is given, that the density of a metal is 9.80 g/ml.

Let the mass of a sample of metal be x.

The sample of metal is dropped in 28.9 ml of water, due to which the volume of the water increases to 56.4 ml.

In order to calculate the mass of a metal, there is a need to use the formula, mass = density * volume

Mass = (9.80 g/ml) (56.4 ml - 28.9 ml)

= (9.80 g/ml) (27.5 ml)

= 2.70 × 10² g or 270 g

6 0

3 years ago

Calculate the [ OH − ] and the pH of a solution with an [ H + ] = 0.090 M at 25 °C . [ OH − ] = M pH = Calculate the [ H + ] and

Valentin [98]

Answer: a) $[OH^-]=1.09\times 10^{-13}$ and pH = 1.04

b) $[H^+]=1.02\times 10^{-11}$ and $pH=10.99$

c) $[H^+]=7.08\times 10^{-11}$ and $[OH^-]=1.41\times 10^{-4}$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-\log [OH^-]$

$pH+pOH=14$

a) $[H^+]=0.090M$

$pH=-\log [0.090]=1.04$

$pOH=14-1.04=12.96$

$12.96=-log[OH^-]$

$[OH^-]=1.09\times 10^{-13}$

b) $[OH^-]=0.00098M$

$pOH=-\log [0.00098]=3.01$

$pH=14-3.01=10.99$

$10.99=-log[H^+]$

$[H^+]=1.02\times 10^{-11}$

c) $pH=10.15$

$10.15=-\log [H^+]$

$[H^+]=7.08\times 10^{-11}$

$pOH=14-10.15=3.85$

$3.85=-log[OH^-]$

$[OH^-]=1.41\times 10^{-4}$

3 0

3 years ago

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]

Answer: The final temperature of the mixture is 49°C

Explanation:

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

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

For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

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

For hot water:

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

2 A certain gas of 25 g at 25°c and 0.65 atm occupies a volume of 23.52L Determine the molecule mass of the gas.

denpristay [2]

Answer:

${ \bf{PV= \frac{m}{M} RT}} \\ \\ { \tt{(0.65 \times 23.52) = \frac{25}{M} \times 0.081 \times (25 + 273)}} \\ \\ M = 39.5 \: g$

8 0

3 years ago