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3 years ago

How can you tell if the content on a website is modified regularly?

Chemistry

2 answers:

just olya [345]3 years ago

5 0

It looks different? That’s the only way unless it says it was edited.

Natalija [7]3 years ago

4 0

A way you would be able to tell if a website is modified regularly is by looking for if the website is open to the public to write on.

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Given the following reaction:

fredd [130]

Answer:D. In order to restore equilibrium, the reaction shifts right, towards products

Explanation:

7 0

2 years ago

How does temperature, pressure and surface area affect the dissolving process

larisa [96]

If I am correct, yes. As I was told in chemistry, the surface area affects the dissolving of the "sugar". If you put regular Surat in a hot cup of coffee, it will dissolve at a quick pace, but what if you put the same amount of sugar in the same amour of coffee, but the sugar was fine powder? It would dissolve even faster since it has more surface area. So temperature does affect the dissolving. Hope this helps!

8 0

3 years ago

Read 2 more answers

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid 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 solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C

7 0

3 years ago

Vishwanath had some money. he spent 3 upon 4 part of money to buy goods for his birthday,1 upon 5 part of money give to his sist

lord [1]

Answer:

The correct answer is - 800.

Explanation:

Given:

Total amount = ? or assume x

spend in buying birthday item = 3/4 of x

given to sister = 1/5 of x

remaining to mother = 40

solution:

the remaning amount = x- (3x/4+x/5) = 4=

=> x- 19x/20 = 40

=> x = 20*40

=> x = 800

thus, the correct answer is = 800

8 0

3 years ago

If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol rema

SVETLANKA909090 [29]

Explanation:

Let us assume that the given data is as follows.

V = 3.10 L, T = $19^{o}C$ = (19 + 273)K = 292 K

P = 40 torr (1 atm = 760 torr)

So, P = $\frac{40 torr}{760 torr} \times 1 atm$

= 0.053 atm

n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

PV = nRT

$0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K$

0.1643 = $n \times 23.97$

n = $6.85 \times 10^{-3}$

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

$6.85 \times 10^{-3} = \frac{mass}{46 g/mol}$

mass = $315.1 \times 10^{-3}$ g

= 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.

4 0

3 years ago