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vova2212 [387]
3 years ago
14

Sorry if it’s not a good pic but...What’s the answer?!?

Chemistry
2 answers:
solniwko [45]3 years ago
8 0
The answer is A! Have a good day! :))
arsen [322]3 years ago
6 0

Answer:

If the products contain 3 nitrogen atoms, then so did the reactants since overall mass is conserved in a chemical reaction.

Explanation:

so the answer is A

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CAN ANYONE SHARE 10 MEMES AND BAG THE BRAINLISST
Semmy [17]
*clink* a toast *clink* a toast *clink* a toast *clink* *clink*

5 0
3 years ago
Read 2 more answers
While in Europe, if you drive 111 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite
kifflom [539]
First put everything into the same measurements.
111 km = 111km
42mpg = 67.592kpg
1 U.K. Gallon = 4.546 liters
67.592/4.546 = 14.8684557853
42mpg = 14.87kpl (kilometers per liter)
1.10 euros/liter = 1.10 euros/liter

If you drive 111 km a day and go 14.87 km a liter, divide 111 by 14.87 to get about 7.46, which is how many liters you will use a day. Multiply the price, 1.10 euros, by the number of liters, 7.46, to get the daily price of gas, which is about 8.206 a day. Multiply by 7 (num of days in week) to get about 57.44

It cost about 57.44 euros for one week of gas.

I hope I'm not too late! And hopefully that is right! No guarantees lol
5 0
3 years ago
What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific he
atroni [7]
LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.
4 0
4 years ago
er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The
liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0
4 years ago
Given that the Ksp value for Ca3(PO4)2 is 8.6×10−19, if the concentration of Ca2+ in solution is 4.9×10−5 M, the concentration o
nasty-shy [4]

Answer:

.0027 M

Explanation:

We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:

Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)

Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2

[PO3−4]=0.0027 M

5 0
3 years ago
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