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iren2701 [21]
2 years ago
13

Chrissy is doing her chemistry homework and comes across the chemical formula for 3 molecules of nitric acid, represented by 3HN

O3.
Which of the following shows the correct number of atoms for Chrissy's homework problem?
A-
3 hydrogen
1 nitrogen
3 oxygen

B-
1 hydrogen
I nitrogen
3 oxygen
C-
3 hydrogen
3 nitrogen
9 oxygen
Chemistry
1 answer:
dybincka [34]2 years ago
5 0

Answer:

C

Explanation:

The 3 goes to every term in the molecule

3 NHO3

So its

3x1 N's

3x1 H's

3x3 O's

You might be interested in
7.02 x 10^23 molecules of X2, a ssubstance consisting of diatomic molecules, has a mass of 296 grams. Determine the atomic weigh
Fiesta28 [93]

Answer:

d. 127 g/mol.

Explanation:

Hello!

In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:

mol=7.02x10^{23}molec*\frac{1mol}{6.022x10^{23}molec}=1.17mol

Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:

MM=\frac{296g}{1.17mol}\\\\MM=254g/mol

It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.

Best regards!

4 0
2 years ago
The combination of potassium-sparing diuretics and salt substitutes can result in dangerously high blood levels of:
alisha [4.7K]

Answer:

b. potassium.  

Explanation:

Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.

Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.

4 0
3 years ago
Scientist name: ______________
nikitadnepr [17]
This is easy… like you can’t take 20 min to search this up in Googl
5 0
2 years ago
Read 2 more answers
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
2 years ago
Why aren't flowers sold at a monastery math worksheet answers?
maria [59]

1) x/9 + 16 = 5.

a) x/9 + 16 - 16 = 5 -16; subtraction.

b) x/9 = -11; simplifying.

c) x/9 · 9 = -11 · 9; multiplication.

d) x = -99; simplifying.

2) 5 · (2a - 3) = 60.

a) 10a - 15 = 60; distributive.

b) 10 - 15 + 15 = 60 + 15; addition.

c) 10a = 75; simplifying.

d) 10a ÷ 10 = 75 ÷ 10; commutative.

e) a = 7.5; simplifying.

3) 4 n + (7 + n) = 52.

a) 4n + (n + 7) = 52; commutative.

b) ( 4n + n) + 7 = 52; associative.

c) 5n + 7 = 52; combining.

d) 5n + 7 - 7 = 52 - 7; subtraction.

e) 5n = 45; simplifying.

f) 5n ÷ 5 = 45 ÷ 5; division.

g) n = 9; simplifying.

4) 8t - 3 · (4t + 2) = 16.

a) 8t - 12 t - 6 = 16; distributive.

b) -4 t - 6 = 16; combining.

c) - 4 t - 6 + 6 = 16 + 6; addition.

d) -4t = 22; simplifying.

e) -4t ÷ (-4) = 22 ÷ (-4); division.

f) t = -5.5; simplifying.

This is the question I found on internet.

8 0
3 years ago
Read 2 more answers
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