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2 years ago

How long is a bench? Select the best estimate.

Chemistry

1 answer:

Alenkinab [10]2 years ago

8 0

Answer:

4 meters

Explanation:

4 centimeters and millimeters are too small, while 4 kilometers is too large.

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Which of the metals below will dissolve in hydrochloric acid, HCl? gold, Au copper, Cu silver, Ag magnesium, Mg

Lera25 [3.4K]

Magnesium (Mg)

The reason for this is the reactivity of the listed metals. Gold and silver are extremely unreactive metals. It is because of this unreactive nature that they remain in good condition for long periods of time, and are preferred in jewelry. Copper, although more reactive than gold and silver, is still not reactive enough to react with HCl.
The only metal that will react is magnesium.

7 0

3 years ago

What mass of NaCl would you have dilute to 1L to make up a 0.25M solution?<br> Asap please

gavmur [86]

Answer: 0.25m

Im not sure but i think

Explanation:0.25/0.25

3 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

Carbon forms a great variety of stable compounds. Which of the following is not a reason for this variety?

katen-ka-za [31]

Answer:

Explanation:

Carbon atoms are not strongly electronegative and tend to form covalent bonds.

4 0

3 years ago

If your front lawn is 18.0 feet wide and 20.0 feet long. And each square foot of lawn accumulates 1450 new snow flakes every min

creativ13 [48]

To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.

Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:

1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour

It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.

6 0

3 years ago