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3 years ago

Which of the following pairs of elements is most likely to form an ionic compound?

Chemistry

1 answer:

Scrat [10]3 years ago

5 0

Explanation:

magnesium and sodium because they are minerals and part of iron

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Water has which of the following IMFs present between molecules?

BlackZzzverrR [31]

Answer:

Water has strong hydrogen bond dipole-dipole intermolecular forces that give water a high surface tension and a high heat of vaporization and that make it a strong solvent.

Explanation:

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3 years ago

A blank is a coil of wire that produces a magnetic field when it carries an electric current

Lostsunrise [7]

A spring - also an electromagnet. Also know as a slinky!

7 0

4 years ago

Read 2 more answers

Help on question 1b please? (image attached)

aliya0001 [1]

I believe your answer Is C. An ammonia molecule has a trigonometrical pyramidal shape. Figure C has a <span>has a trigonometrical pyramidal shape.</span>

I hope I help

3 0

3 years ago

How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102

Ede4ka [16]

Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles$

1 mole of hydrogen $(H_2)$ = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms

17.5 mole of hydrogen $(H_2)$ = $\frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25}$ atoms

There are $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0

3 years ago

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx

Anna [14]

Answer:

$M=235.42g/mol$

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

$2n_{acid}=n_{base}$

It means that the moles of acid can be computed given the volume and concentration of NaOH:

$n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol$

It means that the approximate molar mass of the acid is:

$M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol$

Best regards!

5 0

3 years ago