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3 years ago

ASAP MULTIPLE CHOICE WILL MARK BRAINLIEST

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

7 0

Answer:

Atoms consist of at least some negatively charged particles (Electron)

Explanation:

J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.

Please correct me if I'm wrong.

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In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch

siniylev [52]

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0 μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0 μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7 μg

3 0

3 years ago

A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold

Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

$\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}$

3 0

3 years ago

Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h

balu736 [363]

CxHy + O2 --> x CO2 + y/2 H2O

Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound

Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound

Find the mass of C and H in the compound:

.430mol x 12 = 5.16 g C

.166mol x 1g = .166g H

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0

4 years ago

How many moles of KNO3 are in 500 mL of 2.0 M KNO3? ___ mol KNO3

weeeeeb [17]

Answer: 1mole

Explanation:

Mole = concentration× Volume (dm3)

Mole = 2× 500/1000

4 0

3 years ago

Read 2 more answers

How many kilograms of oseltamivir would be needed to treat all the people in a city with a population of 400000 people if each p

madam [21]

Answer:

300 kg

Explanation:

The number of people in the city, p = 400,000

The number of capsules of oseltamivir each person consumes per day, n = 2 capsules

The number of days each person consumes the oseltamivir capsules, t = 5 days

The mass of oseltamivir in each capsule, m = 75 mg

The mass of oseltamivir needed to treat all the people in the city, M, is given as follows;

M = n·t·m·p

∴ M (in milligrams) = (2 × 5 × 75 × 400,000) mg = 300,000,000 mg

1,000,000 mg = 1 kg

∴ 300,000,000 mg = 300 kg

The mass of oseltamivir needed to treat all the people in the city, M = 300 kg

3 0

3 years ago