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frez [133]
3 years ago
15

Not sure what to do can you help

Chemistry
1 answer:
NeTakaya3 years ago
7 0

sorry i don't know the answer i'm really sorry

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The number of Joules needed to raise the temperature of 10 grams of water from 20°C to 30°C is
e-lub [12.9K]

Answer:

418 Joules

Explanation:

Q=10(4.18)(10)

6 0
3 years ago
Calculate the number of molecules in 64.5 grams of Nitrogen monoxide, NO
Alborosie

N=9.933

Explanation:

no of mole = m÷mr

then ,no of molecules=n×la

3 0
3 years ago
Calculate the number of c h and o atoms in 1.50g of glucose (c6h12o6) a sugar
goldfiish [28.3K]
Total number of atoms is 5.016×10²¹ Carbon and oxygen states 3.01×10²³ respectively, while hydrogen 6.02×10²³
8 0
4 years ago
if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.

4 0
3 years ago
The second atomic structure YES <br> No
gizmo_the_mogwai [7]

Answer

The second part of the theory says all atoms of a given element are identical in mass and properties. The third part says compounds are combinations of two or more different types of atoms. The fourth part of the theory states that a chemical reaction is a rearrangement of atoms.

Explanation:

4 0
3 years ago
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