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3 years ago

Is francium cation or anion?

Chemistry

1 answer:

lyudmila [28]3 years ago

3 0

Answer:

Cation.

Explanation:

This is due to Francium being a positively charged element. It's in the first period, which means it has one extra electron, making it a positively charged element.

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Balance the following chemical reaction: _NaOH + _H2SO4 ----->_Na2SO4 +_H20

Dima020 [189]

Answer:

Sodium Hydroxide + Sulfuric Acid = Sodium Sulfate + Water

2NaOH + H2SO4 → Na2SO4 + 2H2O

Explanation:

To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.

Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.

Ionic charges are not yet supported and will be ignored.

Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.

Compound states [like (s) (aq) or (g)] are not required.

You can use parenthesis () or brackets [].

8 0

2 years ago

How many hybrid orbitals do we use to describe each molecule c2h5no (4 c−h bonds and one o−h) key?

yarga [219]

Answer:

see attached

Explanation:

6 0

3 years ago

the problem say what is the volume (in kL) of 3.7505 x 10^4 mg of iron? Iron has a density of 7.87 g/ml. Use correct significant

Elis [28]

So first find the volume

so 7.78g/ml and (3.7505 times 10^4) grams
therefor we dividde (3.7505 times 10^4) by 7.78 and get how many ml

the answer is 4820.69 ml

kL means kilo lieters
kilo=1000
kL=1000Liters
ml=milileters=1/1000 leiter
1000ml=1L
therfor
1kL=1000L
1000L=1000 times (1000ml)
1000L=1,000,000ml
1kL=1,000,000ml

so to convert to kL divide 4820.69 by 1,000,000
0.00482069
convert to scientifiic notation
4.8 times 10^-3

the answer is 4.8 times 10^-3 kL

4 0

2 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 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

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

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

2 years ago

What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.

pav-90 [236]

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if 12.41 g of MgBr₂ are dissolved in 2.55 L of water

then X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in 4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

7 0

3 years ago