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kirill [66]
3 years ago
6

Can someone please look at this?

Chemistry
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

\large \boxed{\text{110 mmHg}}}

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases exerts its pressure separately from the other gases.

In other words, if a gas makes up 3.0 % of the atmosphere, its partial pressure is 3.0 % of the total pressure.

p_{\text{N}_{2}} = p_{\text{tot}} \times \dfrac{\text{3.0 parts N}_{2}}{\text{100 parts atmosphere}}= \text{3700 mmHg} \times \dfrac{\text{3.0 mmHg}}{\text{100 mmHg}} = \textbf{110 mmHg}\\\\p_{\text{N}_{2}} = \large \boxed{\textbf{110 mmHg}}

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In one experiment, magnesium metal melts. In a second experiment, magnesium metal ignites as it combines with oxygen. Classify t
joja [24]

Answer:

magnesium metal melts = physical change

magnesium metal ignites = chemical change

Explanation:

<em>Physical changes</em> are those in which the identity of the subtance <u>remains unaltered</u>. No new compounds are formed. They involve generally changes in <u>agreggation states of matter</u>: solid, liquid or gas. The first experiment, in which magnesium metal melts is a physical change because it only changes the state of matter, from solid to liquid, but it is still magnesium metal.

Conversely, <em>chemical changes</em> involve atoms combinations to form new compounds. The second experiment, in which magnesium metal ignites, is a chemical change. After the change, magnesium metal is no longer the metal but a metal oxide.

8 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp
bija089 [108]
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
5 0
3 years ago
A scientist wants to determine which fertilizer is more effective—Fertilizer X or Fertilizer Y. The best way for her to proceed
Alecsey [184]

Answer:

The correct answer is "three groups of plants—a group fertilized by X, a group fertilized by Y, and a control group with no fertilizer".

Explanation:

I had to look for the problem to know the options.

The best way to determine which fertilizer is most effective is to have three evaluation groups. One group will be tested with fertilizer X and another with fertilizer Y. By leaving the third group without applying fertilizer, this will be the general pattern for comparing the effectiveness of the other two.

Have a nice day!

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