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Anarel [89]
3 years ago
15

QUESTION 5

Chemistry
1 answer:
amid [387]3 years ago
8 0

Answer:

m+/+

commen+/0

para +/-

compe-/-

Explanation:

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What object has a mass of 10 grams and a volume of 2 mL
Elena-2011 [213]

Answer:

Destiny = 5 g/ml

Explanation:

The formula for density is d=

Knowing this, divide your mass by volume:

10 g / 2 ml = 5 g/ml

Final answer:

Destiny = 5 g/ml

3 0
2 years ago
The noble gases are the least
cricket20 [7]

Answer:

reactive nonmetals since they have a full valence shell (that's why they're stable).

4 0
3 years ago
Calculate the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remain
Reil [10]

The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:

3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂

This is a redox reaction, which <u>is a chemical reaction in which one or more electrons are transferred between the reagents</u>, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.

<u>The mass of chromium metal produced in the above reaction will be,</u>

425.0 mL x \frac{1 L}{1000 mL} x  \frac{0.25 mol Cr(NO_{3})_{3}  }{1 L} x \frac{2 mol Cr  }{2 mol Cr(NO_{3})_{3} } x \frac{51.9961 g Cr}{1 mol Cr} = 5.52 g

So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.

6 0
3 years ago
How many grams of NaHCO3are in 3.75 x 10 ^-3 moles of NaHCO3<br><br> PLS HELP DUE TODAY
hodyreva [135]

Answer: m= 3.15x10-3 g NaHCO3

Explanation: To find the mass of NaHCO3 we will use the relationship between moles and molar mass. The molar mass of NaHCO3 is 84 g.

3.75x10-5 moles NaHCO3 x 84 g NaHCO3 / 1 mole NaHCO3

= 3.15x10-3 g NaHCO3

3 0
3 years ago
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

Best regards.

8 0
3 years ago
Read 2 more answers
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