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3 years ago

Please help this assignment is due today. First person to answer gets 17 points. Please help ASAP.

Chemistry

1 answer:

DochEvi [55]3 years ago

5 0

Answer:

hi umm can you please be my friend please thankyou so much and the answer for this is can i please get brainlist and can you please like this and rate it so now the answer is what type of chemical bond commonly forms between adjacent water molecules? of ionic, covalent, and hydrogen bonds, ionic bonds are the most common type of chemical bonds found in the body.please may i grt brain-list

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What is the % composition of Carbon in Chromium (iii) Carbonate

photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

$Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3$

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

$\begin{gathered} C\rightarrow3\times12=36 \\ \\ O\rightarrow9\times16=144 \\ \\ Cr\rightarrow2\times52=104 \end{gathered}$

The molar mass will be thus:

$M=36+104+144=284\text{ g/mol}$

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

$\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\ \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}$

The percent composition of Carbon is thus 12.7 %.

8 0

1 year ago

A 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H 2 SO 4 , and the hydrogen gas produced

Rama09 [41]

Answer:

0.981atm

Explanation:

According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.

total pressure= 1.01at

Number of gases=2

Gases: water vapor and hydrogen

partial pressure of water vapor= 0.029atm

1.01= partial pressure of water vapor+ partial pressure of hydrogen

1.01= 0.029 + partial pressure of hydrogen

partial pressure of hydrogen = 0.981atm

6 0

3 years ago

Give an example of each of the following chemical changes.

Andrews [41]

Answer:

Give an example of each of the following chemical changes.

(a) A photochemical reaction involving

(i) silver salt

(ii) water

(b) A reaction involving

(i) blue solution

(ii) formation of dirty green precipitate

(d) A reaction where colour change is noticed.

Explanation:

a) A photochemical reaction involving silver salt is used in black and white photography.

AgCl breaks down and converts into Ag during this photochemical reaction.

(ii) Photochemicalreaction involving water takes place in plants during the photosynthesis process.

Plants prepare food(carbohydrate) by using sunlight water and CO2 gas.

(b) A reaction involving

(i) blue solution:

For example reaction of copper sulfate solution with an iron nail.

When an iron nail is placed in CuSO4 blue color solution, then it changes to green color and reddish-brown solid deposits at the bottom of the container.

(ii) Reaction of ferrous sulfate with NaOH forms a dirty green precipitate of ferrous hydroxide.

For example when HCl gas reacts with ammonia gas, then a white solid of ammonium chloride will be formed.

(d) A reaction where the color change is noticed.:

When an iron nail is placed in CuSO4 blue color solution, then it changes to green color and reddish-brown solid deposits at the bottom of the container.

7 0

3 years ago

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of

Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the alloy is 2.324 J/g°C

3 0

3 years ago

Why it is so hard to find life on other planets.

STatiana [176]

Answer:

Because people on earth may not have the correct technology.

Explanation:

5 0

3 years ago