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3 years ago

Balancing equations

Chemistry

1 answer:

vovangra [49]3 years ago

8 0

Answer:

4. 1HCL + 1NAOH- I NACL+1H2O

Explanation:

there is 2 hydrogens on the left side of the equation, balancing out to the other side of the equation. everything else would be balanced out too

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1. 2.0 mL/s x 4.0 hrs = ? L How do you do this conversion problem

Phantasy [73]

Answer:

29 L.

Explanation:

Hello!

In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:

$2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L$

Which turns out 29 L with 2 significant figures.

Best regards!

6 0

3 years ago

Is honey homogeneous or heterogeneous

RideAnS [48]

Honey is a homogeneous mixture because it has the properties that definehomogeneous solutions or mixtures.

7 0

3 years ago

Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was react

stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of Al produce 2 moles of Al₂O₃

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

8 0

3 years ago

What element are usually shiny , can be bent or stretched and conduct electricity

Alik [6]

Answer:

gold and copper

Explanation:

but I think there is 1 more

8 0

3 years ago

Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu

Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. So there are 7 valence electrons.

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are 4 valence electrons.

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are 3 valence electrons.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are 8 valence electrons.

Therefore, this element belongs to the group 8 of the periodic table.

3 0

3 years ago