An aqueous solution of 6.70 M ammonia, NH3, has a density of 0.950 g/mL. The percent by mass of NH3 in the solution is %.

The given question is incomplete. The complete question is:

A chemist adds 0.85 L of a 0.0050M calcium sulfate to a reaction flask. Calculate the mass in grams of calcium sulfate the chemist has added to the flask. Round your answer to significant digits.

Answer: The mass in grams of calcium sulfate the chemist has added to the flask is 0.58 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in L

moles of [tex]CaSO_4[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{136g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.0050=\frac{x}{136\times 0.85}[/tex]

[tex]x=0.58 g[/tex]

Therefore, the mass in grams of calcium sulfate the chemist has added to the flask is 0.58 g

Answer:

try looking the equations or find a calulator for those type of eqautions

Explanation:

Answer: You pedaling the bike is transforming chemical energy

Explanation:

Answer:

B. a chemical reactions that occurs in the chloroplast of a plant.

Explanation:

Answer:

B.

Explanation:

it is b because I took science

lol have a good day

Answer:

I don't think that answer is right. I got 543.04 grams.

Explanation:

I think the question is asking for grams in the samlpe instead of the moles in the sample because it already gives you the amount of moles.

Answer:

A) HCI( hydrochloric acid)

Explanation:

HCI is a strong acid because it has more number of hydrogen ions whereas acetic acid contains less number of hydrogen ions so its a weak acid

Answer:

See explanation

Explanation:

An initiation step is the first step in a free radical mechanism. In this step, free radicals are generated from a stable compound. The initiation step tends to increase the concentration of radicals present in the system.  It leads to a net increase in the number of free radicals in the system.

For the chlorination of 2,2-dimethylpropane, the initiation step is;

Cl2 + hv --------->2Cl.

Note that hv is used to represent light

Answer: B

Mutualism.

Explanation:

Answer:

0.10M of Ba²⁺ is the concentration of the metal in excess

Explanation:

Based on the chemical reaction:

K₂CO₃(aq) + Ba(NO₃)₂(aq) → BaCO₃(s) + 2KNO₃(aq)

1 mole of potassium carbonate reacts per mole of barium nitrate

To solve this question we need to find the moles of each salt to find then the moles of Barium in excess:

Moles K₂CO₃:

0.025L * (0.25mol / L) = 0.00625moles K₂CO₃ = moles CO₃²⁻

Moles Ba(NO₃)₂:

0.030L * (0.40mol/L) = 0.012 moles of Ba(NO₃)₂ = 0.012 moles of Ba²⁺

That means moles of Ba²⁺ that don't react are:

0.012 mol - 0.00625mol = 0.00575 moles Ba²⁺

In 25 + 30mL = 55mL:

0.00575 moles Ba²⁺ / 0.055L =

The concentration of the excess metal ion after the precipitation reaction is complete is 0.1045 M

From the question,

We are to determine the concentration of the excess metal ion after the precipitation reaction is complete

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

K₂CO₃ + Ba(NO₃)₂ → BaCO₃ + 2KNO₃

This means

1  mole of K₂CO₃ is required to react completely with 1 mole of Ba(NO₃)₂

Now, we will determine the number of moles of each reactant present.

Volume = 25.0 mL = 0.025 L

Concentration = 0.25 M

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of K₂CO₃ present = 0.25 × 0.025

Number of moles of K₂CO₃ present = 0.00625 mole

Volume = 30.0 mL = 0.03 L

Concentration = 0.40 M

∴ Number of moles of Ba(NO₃)₂ present = 0.40 × 0.03

Number of moles of Ba(NO₃)₂ present = 0.012 mole

Since

1 mole of K₂CO₃ is required to react completely with 1 mole of Ba(NO₃)₂

Then,

The 0.00625 mole of K₂CO₃ will react with 0.00625 mole of Ba(NO₃)₂

∴ Only 0.00625 mole of the Ba(NO₃)₂ will react

This means Ba(NO₃)₂ is the excess reactant.

Now, we will determine the number of moles of Ba(NO₃)₂ remaining

Number of moles of Ba(NO₃)₂ remaining = 0.012 mole - 0.00625 mole

Number of moles of Ba(NO₃)₂ remaining = 0.00575 mole

Now, for the concentration

Using the formula

[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]

Volume = Total volume of the solution = 0.025L + 0.03L

Volume = 0.055 L

∴ Concentration of the excess metal ion = [tex]\frac{0.00575}{0.055}[/tex]

Concentration of the excess metal ion = 0.1045 M

Hence, the concentration of the excess metal ion after the precipitation reaction is complete is 0.1045 M

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Answer:

Alkene

Explanation:

The IR spectrum of an organic compound shows us the functional groups present in the compound.

By looking at the various stretching frequencies given in wave numbers, we can decide what functional groups are in the compound and hence the class of organic compounds to which it belongs.

The stretch 3080 cm-1 shows a mono-substituted alkene

The stretch 2950 cm-1 shows a C-H bond

The stretch 1620 cm-1 shows an alkenyl stretch.

Hence the compound is an alkene.

Answer:

Explanation:

Airs and trees

Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

Answer:

[tex]r=0.0341\frac{M}{s}[/tex]

Explanation:

Hello!

In this case, considering the given table, we are able to represent the symbolic rate law as shown below:

[tex]r=k[BF_3]^m[NH_3]^n[/tex]

Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):

- Experiment 1 and 2 for the calculation of n:

[tex]\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1[/tex]

So the order of reaction with respect to NH3 is 1.

- Experiment 3 and 4 for the calculation of m

[tex]\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1[/tex]

So the order of reaction with respect to BF3 is also 1.

Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:

[tex]k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}[/tex]

Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:

[tex]r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}[/tex]

Best regards!

4 cm³

Math

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Chemistry

Gas Laws

Density = Mass over Volume

Step 1: Define

D = 25 g/cm³

m = 100 g

Step 2: Solve for V

Answer:

false

Explanation:

Answer:

its a diffrent color :)

Explanation:

Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.

Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.

Answer:

Cccccccccccccccccccccccccccccccccccccccccccc

Answer:

The answer is B (to transport nutrients and waste throughout the body)

Explanation:

Took the test

Answer:

The molar concentration of normal saline is 0.15 mol/L.

Explanation:

The mass of NaCl in 0.85% of normal saline is:

[tex] 0.85 \% = \frac{0.85 g}{100 ml} [/tex]

[tex] m_{NaCl} = 0.85 g [/tex]

The molar concentration can be found using the following equation:

[tex] C = \frac{n}{V} [/tex]        

So we need to find the number of moles of NaCl in 1 liter of solution.

[tex]m_{NaCl} = \frac{0.85 g}{100 mL}*\frac{1000 mL}{1 L}*1 L = 8.5 g[/tex]

[tex]n = \frac{m_{NaCl}}{M}[/tex]

Where:

M: is the molar mass of NaCl = 58.44 g/mol

[tex] n = \frac{8.5 g}{58.44 g/mol} = 0.15 moles [/tex]

Hence, the molar concentration is:

[tex]C = \frac{0.15 moles}{1 L} = 0.15 mol/L[/tex]        

Therefore, the molar concentration of normal saline is 0.15 mole per liter.

I hope it helps you!      

Explanation:

I mol of water = 22.4 lts of O2

thus, 4.21 moles of water = 4.21× 22.4= 94.304 lts.

The locations with subatomic particles with the least mass are locations 1 and 4.

An atom is usually divided into two main regions:

The three sub-particles previously mentioned (neutrons, protons, and electrons) differ not only in their location but also in their mass. Indeed, electrons have a lower mass if compared to neutrons and protons.

Based on this, the outer region (locations 1 and 4) has sub particles with the least mass.

Learn more about atom in: https://brainly.com/question/13981855

Answer:

1 and 4

Explanation:

took the test

Just saying thank you person above you are very great and smart. You deserve a hugeeeeeeee thank you. So, thank you!!!!!!!!!

290 g Na₂O

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

4.7 mol Na₂O

Step 2: Identify Conversions

Molar Mass of Na - 22.99 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Na₂O - 2(22.99) + 16.00 = 61.98 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

291.306 g Na₂O ≈ 290 g Na₂O

Answer:

[tex]pH=4.63[/tex]

Explanation:

Hello!

In this case, since the ionization of ammonia, which is a weak base, is written as:

[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]

We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:

[tex]pH=pKb+log(\frac{[conj\ acid]}{[base]} )[/tex]

In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:

[tex]pH=4.75+log(\frac{0.15}{0.2} )\\\\pH=4.63[/tex]

Best regards!

H₂SO₄ is composed of the most atoms

Given

Compounds of A. H2SO4 B. Li2N C. Mg(OH)2 D. NaF

Required

The most atoms

Solution

Compounds are composed of atoms of the same or different types

The number of atoms can be indicated from the subscript of each of the constituent atoms of the compound

A. H₂SO₄ : 2 atoms H, 1 atom S, 4 atoms O = 7 atoms

B. Li₂N : 2 atoms L, 1 atoms N = 3 atoms

C. Mg(OH)₂ : 1 atom Mg, 2 atoms O, 2 atoms H = 5 atoms

D. NaF : 1 atom Na, 1 atom F = 2 atoms

Answer:

See explanation.

Explanation:

Hello!

In this case, since we are asked to note down the complete molecular equation, complete ionic equation and the net ionic equation for the reaction between acetic acid (weak acid) and barium hydroxide (strong base), we proceed as shown below:

- Complete molecular equation:

[tex]2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)[/tex]

- Complete ionic equation: in this case, since acetic acid is a weak one, it is not ionized, so we do this:

[tex]2CH_3COOH(aq)+Ba^{2+}(aq)+2OH^-(aq)\rightarrow Ba^{2+}(aq)+2CH_3COO^-(aq)+2H_2O(l)[/tex]

- Net ionic equation: in this case, we cancel out the barium ions as they are the spectators one because they are present at both reactants and products:

[tex]2CH_3COOH(aq)+2OH^-(aq)\rightarrow 2CH_3COO^-(aq)+2H_2O(l)\\\\CH_3COOH(aq)+OH^-(aq)\rightarrow CH_3COO^-(aq)+H_2O(l)[/tex]

Best regards!