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Helen [10]
2 years ago
9

When U-238 undergoes radioactive decay by losing an alpha particle, the other product is

Chemistry
2 answers:
lianna [129]2 years ago
6 0
The correct answer is Th-234
myrzilka [38]2 years ago
6 0

TH-234 I took the test

You might be interested in
. In which reaction is nitric acid acting as an oxidising agent?
Talja [164]

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

Regards.

8 0
2 years ago
Read 2 more answers
Mangrove trees are affected by several factors. Which is an abiotic factor contributing to the energy cycle of a mangrove ecosys
kondor19780726 [428]

Answer:

Just think of whats in a forest ecosystem

for example

Water

sunlight

tempreture

4 0
3 years ago
What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol
statuscvo [17]

Answer:

58.92 g EDTA

Explanation:

315.1 mL = .3151 L    

M = Moles / Liter

.3151 L  x <u>0.5 mol EDTA</u>  x  <u>374 g EDTA</u>  =  58.92 g EDTA

                1 L EDTA             1 mol EDTA

4 0
2 years ago
1. Plants that perform photosynthesis convert sun (solar) energy to<br> food energy.
Alinara [238K]

Answer:

true

Explanation:

one reason is because plants need sunlight and water because it is their source of food

3 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
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