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2 years ago

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Which of theCardiorespiratory fitness is most likely to improve due to which of the following types of exercise? A. bicycling B.

weight lifting C. push-ups D. sit-ups Please select the best answer from the choices provided. A B C D

1 answer:

gizmo_the_mogwai [7]2 years ago

7 0

Answer: A (Bicycling)

Explanation:

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A 0.80 L sample of gas has a temperature of 27°C and a pressure of 0.925 atm. How many moles of gas are present?

ololo11 [35]

Answer:

n= 0.03 moles

Explanation:

Using the ideal gas law:

PV=nRT

nRT=PV

n= PV/RT

n: moles

P: pressure in atm

V= volume in L

R= Avogadro's constant = 0.0821

T= Temperature in K => ºC+273.15

n= (0.925 atm)(0.80 L) / (0.0821)(300.15 K)

n= 0.03 moles

6 0

3 years ago

The 85.2-g sample of the compound x4o10 contains 48.0 g of oxygen atoms. what is the molar mass of element x?

Vedmedyk [2.9K]

X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160

so, moles = 85.2 / (4y+160)

Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0

so, 48 = 16 x 10 x [85.2 / (4y+160) ]

Solve the equation to get y.

y = 31

5 0

2 years ago

PLEASE HURRY LIMITED TIME

Stolb23 [73]

3rd one sorry if it’s wrong

3 0

2 years ago

Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample

goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=2:2=1:1$

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=3:3=1:1$

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.

4 0

2 years ago

Are elecronegativity values higher for metals than nonmetals.

Leya [2.2K]

Non-metals, which are organized on the right side of the periodic table, have higher electronegativity values than the metals.

Hope this helped!

3 0

3 years ago