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stiv31 [10]
3 years ago
6

Plz help ASAP it is number 13 y’all have to answer

Chemistry
1 answer:
photoshop1234 [79]3 years ago
4 0

Answer:

A. Moving electrons generate magnetic forces

Explanation:

This is because when electrons or electric current flows in a wire it creates a magnetic field around the wire.

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How many moles of Na₂CO₃ required to create 9.54 liters of a 3.4 M solution
GarryVolchara [31]

Answer:

The answer to your question is 32.44 moles

Explanation:

Data

moles of Na₂CO₃ = ?

volume = 9.54 l

concentration = 3.4 M

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

number of moles = Molarity x volume

Substitution

Number of moles = (3.4)( 9.54)

Simplification

Number of moles = 32.44

3 0
3 years ago
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When magnesium (mg) metal is placed in hydrochloric acid (hcl), it forms magnesium chloride (mgcl2) and a gas. what gas is forme
Delvig [45]
Magnesium in hydrochloric acid forms H2 (hydrogen) gas according to the balanced chemical equation:
Mg + 2HCl→H2 + MgCl2
6 0
3 years ago
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I want to know if it is true or false​
denpristay [2]

Answer:

wheres the question or picture

Explanation:

3 0
3 years ago
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Calculate the percent ionic, the percent covalent, and the bond length (in picometers) of a chemical bond between phosphorus and
kirza4 [7]

Answer:

The correct option is;

4 percent ionic, 96 percent covalent, 222 pm

Explanation:

The parameters given are;

Phosphorus:

Atomic radius = 109 pm

Covalent radius = 106 pm

Ionic radius = 212 pm

Electronegativity of phosphorus = 2.19  

Selenium:

Atomic radius = 122 pm

Covalent radius = 116 pm

Ionic radius = 198 pm

Electronegativity of selenium= 2.55  

The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;

Using Pauling's alternative electronegativity difference method, we have;

\% \, Ionic \ Character = \left [18\times (\bigtriangleup E.N.)^{1.4}  \right ] \%

Where:

Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36

Therefore;

\% \, Ionic \ Character = \left [18\times (0.36)^{1.4}  \right ] \% = 4.3 \%

Hence the percentage ionic character = 4.3% ≈ 4%

the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%

The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;

The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.

The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.

4 0
3 years ago
How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
bezimeni [28]

Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

6 0
3 years ago
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