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3 years ago

Please help I am stuck

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

8 0

The theory of plate tectonics tells exactly where the continents were before Pangaea divided.

I'm not a 100% sure but I feel strong about is

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Predict the products of the thermal decomposition of strontium nitrate and show the reaction as a word equation.

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Explanation:

2SrO + 4NO2 + O. The thermal decomposition of strontium nitrate to produce strontium oxide, nitrogen dioxide and oxygen. This reaction takes place at a temperature of over 570°C

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2 years ago

What do atomic-scale models show you that your observations of properties cannot?

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The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element.

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Formula for hydroselenic acid

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3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

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3 years ago

Will give brainliest if you answer quickly and correctly :)

AnnZ [28]

I think it’s I
Iodine has 53 protons and 74 neutrons

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2 years ago

Read 2 more answers