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Oksi-84 [34.3K]
3 years ago
10

The beta rays emitted from atomic nuclei are _____. an electron a neutron a proton

Chemistry
2 answers:
meriva3 years ago
7 0

Answer: Electrons  I hope this helped! Your fellow Brainly user, GalaxyGamingKitty.

Sliva [168]3 years ago
6 0

The answer I believe is An Electron


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A sample of hydrogen gas will behave most like an ideal gas under the conditions of
elena55 [62]

Answer:

The Answer is gonna be C) high pressure and low temperature

8 0
2 years ago
A compound's molecular formula must always be different than the compound's empirical formula. TRUE FALSE
Juli2301 [7.4K]

Answer: False

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Example: CH_4 has similar molecular formula and empirical formula as the elements are already present in simplest of the ratios.

C_2H_2 has molecular formula of C_2H_2 but CH as the empirical formula.

5 0
2 years ago
Which words in the sentence are the adjective clause? Alessandra, whom everyone calls "Al," is the best tennis player on the tea
Alinara [238K]
The answer is D. because that makes more sense
4 0
3 years ago
A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

8 0
3 years ago
Am I incorrect ?<br> Or nah?
Mademuasel [1]
I think so... I'm currently learning this too but you should be correct
4 0
3 years ago
Read 2 more answers
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