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3 years ago

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If you were asked to convert 25 mg to the unit hg, which of the following would be the first fraction used in the conversion? te

n to the negative third power hg over one g ten to the negative third power mg over one hg ten to the negative third power mg over one g ten to the negative third power g over one mg

1 answer:

NemiM [27]3 years ago

6 0

The conversion factor is added to the original given unit so that you can end up with the final unit. Basically, the conversion unit does not change the value because the factor is just equal to 1. You just manipulate the units by cancelling them.

Among the given choices, the fraction which is equal to 1 is 10⁻³ g/ 1 mg, because there are 1,000 mg per 1 g. In reverse, that would be 10⁻³ of a gram.

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Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

Sphinxa [80]

Answer:

$P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

Explanation:

The first step is:

$P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}$

Second step is:

$P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}$

Multiplying second step by 2, and adding both the steps, we get that:

$P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}$

Cancelling common species, we get that:

$P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}$

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3 years ago

Need help asap!

Scilla [17]

Answer: 1. C. polar covalent: electrons shared between silicon and sulfur but attracted more to the sulfur

2. B) $NH_3$

3. B) Fluorine

Explanation:

1. A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of sulphur- electronegativity of silicon = 2.5 -1.8 = 0.7

Thus as electronegativity difference is less than 1.7 , the cond is polar covalent and as electronegativity of sulphur is more , the electrons will be more towards sulphur.

2. A molecular compound is usually composed of two or more nonmetal elements. Example: $NH_3$

Ionic compound is formed by the transfer of electrons from metals to non metals. Example: $Mg_3N_2$ , $AlCl_3$ and $LiBr$

3. For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here K is having an oxidation state of +1 and as the compound formed is KZ, the oxidation state of non metallic element Z should be -1. Thus the element Z is flourine which exists as diatomic gas $F_2$

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2 years ago

Which of the following combustion reactions is balanced correctly? A. C4H6 + 5.5O2 4CO2 + 3H2O B. C4H6 + 4O2 4CO2 + 3H2O C. C4H6

Airida [17]

You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.

<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O

element reactant side product side

C 4 4
H 6 3*2 = 6
O 5.5 * 2 = 11 4*2 + 3 = 11

Then, this equation is balanced.

</span>Do the same with the other equations if you want to verify that they are not balanced.

Answer: option A.

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3 years ago

Predict the product formed when the compound shown below undergoes a reaction with mcpba in ch2cl2. mcpba is meta-chloroperoxybe

Aleks04 [339]

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3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago