Dumbledore thinks the students are now confident with significant figures; Snape is not so confident of this... Begrudgingly Snape decides to move forward in his lesson. His next lesson is in naming ionic compounds. He presents his students with sodium hydrogen sulfate and potassium permanganate. Which pair of ionic formulas is correct for these two compounds?(A) NaHSO3, KMnO4
(B) NaHSO2, KMnO4
(C) NaHSO4, KMnO4
(D) NaHSO4, KMnO3
(E) NaHSO3, KMnO3

Answers

Answer 1

Answer:

Answer:

C) $NaHSO_(4), KMnO_(4)$

Explanation:

- For the sodium hydrogen sulfate:

The ending -ate of the word sulfate indicates that the compound comes from the ion sulfate that is $SO_(4)^(-)$ , so the compound formed by this ion will be $NaHSO_(4)$

- For the potassium permanganate:

The ending -ate of the word permanganate indicates that the compound comes from the ion $MnO^(-)_(4)$ , so the compound formed by this ion will be $KMnO_(4)$

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What is the total mass (amu) of carbon in each of the following molecules?(a)CH4
(b)CHCL3
(c)C12H10O16
(d)CH3CH2CH2CH2CH3

Answers

The mass of carbon in $\rm CH_4$ is 12.007 amu, $\rm CHCl_3$ is 12.007, $\rm C_1_2H_1_0O_1_6$ is 144.084 amu, and $\rm CH_3CH_2CH_2CH_2CH_3$ is 60.035 amu.

What is the mass of one Carbon atom?

The mass has been given as the sum of the atomic mass unit in the compound. The mass of 1 atom of carbon is 12.007 amu.

The mass of carbon in the following compounds is given as:

$\rm CH_4$

The number of Carbon units = 1

The mass of carbon in compound = 12.007 amu

$\rm CHCl_3$

The number of Carbon units = 1

The mass of carbon in the compound = 12.007 amu

$\rm C_1_2H_1_0O_1_6$

The number of carbon units =12

The mass of carbon in the compound:

$\rm 12\;*\;12.007\;amu\n=144.084\;amu$

$\rm CH_3CH_2CH_2CH_2CH_3$

The number of carbon units = 5

The mass of carbon in the compound:

$\rm 5\;*\;12.007\;amu\n=60.035\;amu$

Learn more about the mass of an atom, here:

brainly.com/question/5566317

Answer:

The atomic mass of carbon (C) is 12.0107 amu, so if you want to calculate the total mass in each molecule, you just need to multiply the number of carbon atoms in the substance by 12.017. In (a) there is one atom of C, (b) have also one atom of C, (c) have 12 atoms of C, and (d) have five atoms of C. Thus, the total mass (amu) of carbon is:

(a) 12.017 amu

(b) 12.017 amu

(d) 60.085 amu

Show the conversion factor from Patolbf/ft2is 0.02089.

Answers

Explanation:

1 Pascal = 1 N/m²

To convert Pa to lbf/ft²

So, the conversion of N to pound force (lbf) is shown below as:

1 N = 0.224809 pound force (lbf)

The conversion of m² to ft² is shown below:

1 m² = 10.7639 ft²

So,

[tex]1\ Pa=\frac {1\ N}{1\ m^2}=\frac {0.224809\ lbf}{10.7639\ ft^2}

1 Pa = 0.02089 lbf / ft²

Hence proved.

How many atoms does 6H2O contain

Answers

Answer:

two atoms of oxygen. For H2O, there is one atom of oxygen and two atoms of hydrogen.

What is the mass in grams of H₂ that can be formed from 54.6 grams of NH₃ in the following reaction?2 NH₃(g) → 3 H₂(g) + N₂(g)

Answers

9.6 grams of H₂ can be formed from 54.6 grams of NH₃ in the following reaction: 2NH₃(g) → 3H₂(g) + N₂(g).

According to this question, the following balanced equation is given: 2NH₃(g) → 3H₂(g) + N₂(g).

First, we convert the mass of ammonia (NH3) to moles as follows:

moles of NH3 = 54.6g ÷ 17g/mol

moles of NH3 = 3.2mol.

If 2 moles of NH3 produces 3 moles of H2.
3.2 moles of NH3 will produce 4.8 moles of H2.

Next, we convert 4.8moles of H2 to mass as follows:

mass of H2 = 4.8 × 2

mass of H2 = 9.6g of H2.

Therefore, 9.6 grams of H₂ can be formed from 54.6 grams of NH₃ in the following reaction: 2NH₃(g) → 3H₂(g) + N₂(g).

Learn more at: brainly.com/question/8732513?referrer=searchResults

Answer : The mass of $H_2$ is, 9.64 grams.

Explanation : Given,

Mass of $NH_3$ = 54.6 g

Molar mass of $NH_3$ = 17 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $NH_3$ .

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=(54.6g)/(17g/mol)=3.21mol$

Now we have to calculate the moles of $H_2$

The balanced chemical equation is:

$2NH_3(g)\rightarrow 3H_2(g)+N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NH_3$ react to give 3 moles of $H_2$

So, 3.21 mole of $NH_3$ react to give $(3)/(2)* 3.21=4.82$ mole of $H_2$

Now we have to calculate the mass of $H_2$

$\text{ Mass of }H_2=\text{ Moles of }H_2* \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(4.82moles)* (2g/mole)=9.64g$

Therefore, the mass of $H_2$ is, 9.64 grams.

calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass

Answers

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water = 0.28 moles (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

How much is a share in namsek worth as a percent of a share in oxd group

Answers

Answer:

10% is a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4

EXPLANATION:give brainliest

The share of NAMSEK in year 4 is around $ shares (as given from the above table). Hence the value of one share will be $2.5/share

The share of ODX Group Inc., in year 4 is around $ shares (as given from the above table). Hence the value of one share is $

Therefore a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4 will be

= 0.1 or 10 %

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