# Write a balanced nuclear equation for the following: The nuclide uranium-238 undergoes alpha emission.

## Answers

Answer 1
Answer:

Answer:

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Energy

Explanation:

Hello,

When uranium undergoes an alpha decay, it's atomic number is reduced by 2 while it atomic mass is reduced by 4 similar to the elements helium (He) and sometimes alpha decay is represented by a helium atom.

Equation of decay

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Energy

When uranium undergoes an alpha decay, it loses an alpha particle similar to a helium atom and forms thorium which has an atomic number of 90 and atomic mass of 234 with the release of energy.

## Related Questions

Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

### Answers

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

### What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

Learn more about percentage mass:

brainly.com/question/3940233

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes  52.94 g = 63.55 g

1 Sulphur  32.07 g =

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

What is the phase of water at 0.25 atm and 0°C?Water
(liquid)
Pressure (atm)
0.5-
0.25
Ice
(solid)
Water vapor
(gas)
0
000
Temperature (°C)
O A. Gas
O B. Solid and gas
O C. Solid and liquid
D. Solid

### Answers

Water is in the solid phase at 0.25 atm and 0°C.

### In what phase is water at 25?

A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state.

### What phase is water in at 0 C?

Under standard atmospheric conditions, water exists as a liquid. But if we lower the temperature below 0 degrees Celsius, or 32 degrees Fahrenheit, water changes its phase into a solid called ice.

Learn more about the solid phase here brainly.com/question/13396621

#SPJ2

It should be a because the temperature and the atm are to low

Identify the characteristics that describe how human proteins are assembled. Check all that apply.

### Answers

Answer:

Hi,

The assembling of proteins starts with the attachment of m RNA to the ribosomes in the cytoplasm. Each of the ribosome read the code in the m RNA from “start” to “ stop” choosing the specific amino acid building block and removing the unwanted growing protein. The ribosome performs this process in 0.02 seconds and with this rate it is possible for the cell to perform assembling of small protein such as insulin.

Best wishes!

Answer:

A, D, E, F

Explanation:

just took it on edge

A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, each containing 100 g of this liquid, would a 85 kg victim need to consume to reach a toxic level of ethylene glycol

### Answers

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) =  0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

### 0.432 drinks are toxic

A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH, CO) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added

### Answers

Answer:

9.09

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Can someone give me brief background info about chromatography experiment?

### Answers

Chromatography is a pretty accurate description of what happens to ink on wet paper, because it literally means "color writing" (from the Greek words chroma and graphe). Really, though, it's a bit of a misnomer because it often doesn't involve color, paper, ink, or writing. Chromatography is actually a way of separating out a mixture of chemicals, which are in gas or liquid form, by letting them creep slowly past another substance, which is typically a liquid or solid. So, with the ink and paper trick for example, we have a liquid (the ink) dissolved in water or another solvent creeping over the surface of a solid (the paper).

The essential thing about chromatography is that we have some mixture in one state of matter (something like a gas or liquid) moving over the surface of something else in another state of matter (a liquid or solid) that stays where it is. The moving substance is called the mobile phase and the substance that stays put is the stationary phase. As the mobile phase moves, it separates out into its components on the stationary phase. We can then identify them one by one.