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kondaur [170]
3 years ago
7

If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater

Chemistry
2 answers:
agasfer [191]3 years ago
8 1

Answer:A. D. Y.

Explanation:

Disjsjjsdd
2 years ago
Ur dumb
Serggg [28]3 years ago
3 0

Answer: 1. ball A 2. their densities are equal 3. ball q 4. ball y

Explanation:

just took the test for them all right

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Uranium has an atomic weight of 238 amu.
shepuryov [24]

Answer:

U= 238g/mol

U2O5= 556g/mol

Explanation:

Since U= 238

O=16

U3O5= 2(238)+3(16)=556g/mol

4 0
3 years ago
Tin is used to coat tin cans for food storage. If the tin is scratched and the iron of the can exposed, will the iron corrode mo
ivann1987 [24]

The scratched tin can with the iron will more rapidly corrode the iron than the tin.

<h3>What is corrosion?</h3>

The corrosion can be given as the process of the oxidation of the metal into the more stable metal oxide. The chemical oxidation of the metal is attained with the surrounding available oxygen or the water vapors.

The reactivity of the Iron for the oxidation is more as compared to the tin from the reactivity series. Therefore if both tin and iron are exposed to corrosion, iron will be more rapidly corroded.

Learn more about corrosion, here:

brainly.com/question/489228

#SPJ1

6 0
2 years ago
True or false atoms with fewer then 4 outer elections lend elections?
Mamont248 [21]
The answer is false ....

3 0
4 years ago
I need this please my brain hurts
GarryVolchara [31]
8.4 grams. I think but I’m not 100% sure
5 0
3 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
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