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3 years ago

HELP????????

Chemistry

2 answers:

nasty-shy [4]3 years ago

8 0

The answer is B. latitude :)

valentina_108 [34]3 years ago

3 0

<span>The most important determining factor in determining climate is latitude
</span><span />

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URGENT:

Ad libitum [116K]

Answer:

The elements that are commonly positive ions are metals. But there are a few gases that can become positively charged by losing electrons.

8 0

2 years ago

In the periodic table, a set of properties repeats from

jok3333 [9.3K]

in the periodic table, a set of properties repeats from row to row

3 0

3 years ago

Read 2 more answers

You need to determine the specific gravity of a sample. After putting the sample on a lab scale, you know it has a mass of 85 gr

BlackZzzverrR [31]

Answer:

Specific gravity of the sample = 8.947

Explanation:

Specific gravity of a substance is defined as the density of that substance divided by the density of water.

Density of water = 1000g/l

Density of substance = mass/volume

= 85/9.5 x 10^-3

= 8947.37 g/l

SG = 8947.37/1000

= 8.947

3 0

2 years ago

A compound is made up of iron and oxygen, only. The ratio of iron ions to oxide ions is 2:3 in this compound. The IUPAC name for

never [62]

Answer: The correct option is 3.

Explanation: We are given a compound which is made up of iron and oxygen only. The ratio of the two are given as:

$\frac{\text{iron ions}}{\text{oxide ions}}=\frac{2}{3}$

This means that, number of iron ions are 2

Number of oxide ions are 3

From the above information, the formula becomes : $Fe_2O_3$

The valency of iron = 3

Valency of oxide = 2

This compound is named as iron (III) oxide.

Hence, the correct option is 3

6 0

3 years ago

Read 2 more answers

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

6 0

2 years ago