What is a radio wave?

To prepare 250mL of calcium chloride solution with a molar concentration of 1.20mol/L, what mass of calcium chloride would be re

e-lub [12.9K]

Answer:

33.30 grams of CaCl2 will be required

Explanation:

Given,

Volume of solution, V= 250 ml

Molarity of solution, M= 1.20 mol/L

Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111

We know,

Required mass, W= SVM/1000

Now,

W = (111 X 250 X 1.20)/1000

= 33300/1000

= 33.30

Therefore, 33.30 grams of Calcium Chloride will be required.

4 0

3 years ago

What is a scientific explanation for a range of hypotheses and observations supported by testing

worty [1.4K]

It verifies the significant relationships of the observed and hypothesized variables in the study.

Furthermore, the authors utilize and make use of the scientific method in order to have clear basis and evidence for their investigations. Research method is always used to answer every scientific inquiry and in gaining evidential data or knowledge. The scientific method has the following process or at least undergoes the process of
1. Observation
2. Hypothesis
3. Experimentation
4. Interpretation of data
5. Evaluating the data
6. Passing and recording the data

These steps are crucial and the empirical data that these scientists obtain are very important to keep that is why research paper, thesis and dissertations exists.

8 0

3 years ago

HELP ITS OVER DUE AND GRADE BOOK ENDS MARCH 5TH

Ymorist [56]

Mutualism- both organisms benefit
Commensalism- one organism benefits while the other neither is harmed or helped
Parasitism- one organism is benefited while the other is harmed

4 P (flea benefits but dog is harmed)
5 M (both get to eat the honey)
6 C (bird gets a place to live without harming or helping the tree)
7 P (lice get a place to live but humans are harmed)
8 M (both are helped)
9 C (flower isn’t harmed or helped but bee is helped)
10 P (the tree is harmed but the mistletoe is benefitted)

4 0

3 years ago

Dalton’s Law CalculationA mixture of H₂, N₂ and Ar gases is present in a steel cylinder. The total pressure within the cylinder

Zolol [24]

Answer:

A) The partial presssure of CO₂ is 167 mm Hg

B) The partial presssure of N₂ is 354 mm Hg

C) The partial presssure of Ar is 235 mm Hg

D) The partial presssure of H₂ is 86 mm Hg

Explanation:

Dalton's law of partial pressures is basically expressed by the following statement:

The total pressure of a mixture is equal to the sum of the partial pressures of its components.

So initially we have:

$P_{T}$ = total presure of the system (675 mm Hg).

$P_{N_2}$ = partial pressure of N₂ (354 mm Hg).

$P_{Ar}$ = partial pressure of Ar (235 mm Hg).

Using Dalton's law we can find the partial pressure of H₂:

$P_{T}=P_{N_2}+P_{Ar}+P_{H_2}$

675 mm Hg=354 mm Hg + 235 mm Hg + $P_{H_2}$

$P_{H_2}$ = 675 mm Hg - 354 mm Hg - 235 mm Hg

$P_{H_2}$ =86 mm Hg

If CO₂ gas is added to the mixture, at constant temperature, and the volume is the same, the difference between the new total pressure and the previous total pressure is equal to the partial pressure of CO₂.

$P_{T}=P_{N_2}+P_{Ar}+P_{H_2}+P_{CO_2}$

842 mm Hg= 354 mm Hg + 235 mm Hg + 86 mm Hg + $P_{CO_2}$

$P_{CO_2}$ = 842 mm Hg - 354 mm Hg - 235 mm Hg - 86 mm Hg

$P_{CO_2}$ = 167 mm Hg

4 0

3 years ago

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

5.

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0

3 years ago