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3 years ago

Use the ruler to determine the length of the line measured to the correct degree of precision. 13.2 cm 13.25 cm 130 cm 132 mm

Chemistry

2 answers:

ipn [44]3 years ago

6 0

Answer: B) 13.25 cm

Explanation: A centimeter ruler is used to measure the lengths of objects and the lengths are recorded to two decimal places.

The given numbers are 13.2 cm, 13.25 cm, 130 cm and 132 cm. 13.2 cm has only one decimal place, 13.25 cm has two decimal places, 130 and 132 cm have zero decimal places.

So, the only correct choice is 13.25 cm as it has two decimal places and so its more precised.

lapo4ka [179]3 years ago

3 0

Your answer would be 13.25cm

Hope that helps!!!

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The first most obvious thing to note is when naming transitional metals, you have to state its charge with roman numerals (except for 1 if I remember correctly). For example, Iron (lll), iron has a charge of 3.

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2 years ago

A(n) _______ is an organic compound in which a carbonyl group is bonded to a nitrogen atom. (2 points)

sineoko [7]

A(n )amide is an organic compound in which a carbonyl group is bonded to a nitrogen atom. This is <span>usually regarded as derivatives of carboxylic acids in which the hydroxyl group has been replaced by an amine or ammonia.</span>

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3 years ago

Read 2 more answers

A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w

Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

$Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)$

$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

$\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol$

4 0

3 years ago

What is the weighted average of a nail in the sample data given?

Natalka [10]

The weighted average of the nail in accordance with the given data is 11.176g.

<h3>How to calculate weighted average?</h3>

Weighted average is an arithmetic mean of values biased according to agreed weightings.

The weighted average of the nail in the image above can be calculated by multiplying the decimal abundance with the mass of the nail, then summed up as follows;

Weighted average = (decimal abundance × mass 1) + (decimal abundance × mass 2)

Weighted average = (0.12 × 3.3) + (0.88 × 12.25)

Weighted average = 0.396 + 10.78

Weighted average = 11.176g

Therefore, 11.176g is the weighted average of the nail

Learn more about weighted average at: brainly.com/question/28042295

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5 0

1 year ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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3 years ago