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3 years ago

I neeeddddd heeeellllllppppp I'll give you brainliest

Chemistry

1 answer:

aev [14]3 years ago

3 0

not double replacement

it will be synthesis

they are fusing the two elements

plz mark brainliest

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Magnesium hydroxide is used as an antacid in milk of magnesia and reacts with hydrogen chlrode in the stomach to form water and

8_murik_8 [283]

Mg(OH)2 +2HCl---> MgCl2+2H2O

3 0

3 years ago

Explain how the equation for GPE = force x distance.

Mkey [24]

GPE is the gravitational Potential Energy which is given by mgh.

mg is the gravitational force(a force) and h is the height of the object(or distance of the object from earth's surface)

SO GPE = mg (force) x h (distance)

4 0

3 years ago

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

Nitroglycerin (c3h5n3o9) is a powerful explosive. its decomposition may be represented by 4c3h5n3o9 → 6n2 12co2 10h2o o2 this re

professor190 [17]

The balanced chemical reaction describing this decomposition is as follows:
4c3h5n3o9 .............> 6N2 + 12CO2 +10H2O + O2

From the periodic table:
mass of oxygen = 16 grams
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of carbon = 12 grams
Therefore:
mass of C3H5N3O9 = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
mass of O2 = 2(16) = 32 grams

From the balanced chemical equation:
4(227) = 908 grams of C3H5N3O9 produce 32 grams of O2. Therefore, to know the amount of oxygen produced from 4.5*10^2 grams C3H5N3O9, all we need to do is cross multiplication as follows:
amount of oxygen = (4.5*10^2*32) / (908) = 15.859 grams

7 0

3 years ago

Read 2 more answers

As you move across the period on the periodic table what pattern can be described about the electron configurations?

xeze [42]

Answer:

Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.

Explanation:

Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell and number of electron increased by 1.

4 0

3 years ago