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guajiro [1.7K]
3 years ago
14

Can someone plss help me asap !!

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
6 0
I think the answer is barium
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How many moles will 1.875×10*4 cm*3 of a substance Y have ?​
Sati [7]

the volume will be 0.84mol of Y

6 0
3 years ago
Read 2 more answers
Identify the spectator ions in the reaction that occurs between aqueous solutions of potassium hydroxide and hydrochloric acid
m_a_m_a [10]

Answer:

The answer is option A.

K+ Cl-

Hope this helps you

8 0
3 years ago
In the following reaction 2C6H6 + 15O2 12CO2 + 6H2O how many grams of oxygen will react with 10.47 grams of benzene (C6H6)?
IRINA_888 [86]
2 C_{6}H_{6}  +  15O_{2} ----->\ \textgreater \  12CO_{2}  +  6H_{2}O




mol of benzene =  \frac{mass}{Mr}
                                = \frac{10.47g}{(6 * 12) (6 * 1) g/mol}
                                = 0.134 mol

mol of oxygen: 
                 ratio of C_{6} H_{6} :  O_{2}
                 =  2 : 15
                 =  1 : 7.5

: . mol of O_{2} = 0.134mol * 7.5
                                         = 1.01 mol

Mass of Oxygen = mol * Mr
                           = 1.01 mol * (16*2) g/mol
                           = 32.22g 

Note: Mr is molar mass
8 0
3 years ago
Please answer this time without just getting points D.​
xz_007 [3.2K]
D is the answer here look

4 0
3 years ago
2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2
Sergeu [11.5K]

Answer: 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles

2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2

According to stoichiometry :

As 1 mole of O_2 is given by = 2 moles of KMnO_4

Thus 0.05 moles of O_2 is given by =\frac{2}{1}\times 0.05=0.10moles  of KMnO_4

Mass of KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g

Thus 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

5 0
3 years ago
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