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3 years ago

A cypress tree is the sprout stage. what would you expect to notice when observing this plant?

Chemistry

1 answer:

Anna007 [38]3 years ago

8 0

Answer:

A mature stem that has thick bark

Explanation:

i dont know but that is what i put on my test but if you had your awnser right tell me

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Pls answer this ASAP thank you

vaieri [72.5K]

Answer:

The anwer is not D the anwer is A

Explanation:

7 0

3 years ago

Bismuth has a density of 9.80 g/cm^ 3 . What is the of 3.74 cm ^ 3 of Bi?

Ierofanga [76]

Explanation:

density = mass/ volume

mass= density × volume

mass= 9.80× 3.74

mass= 36.652 g

3 0

3 years ago

What is the pH of a solution that contains 0.003 M HBr?*<br> O 1.0<br> O 2.5<br> O 5.8<br> O 11.5

maxonik [38]

Answer:

pH = 2.5

Explanation:

Given data:

Concentration of HBr = 0.003 M

pH of solution = ?

Solution:

Formula:

pH = -log[H⁺]

by putting values,

pH = -log [ 0.003]

pH = 2.5

5 0

3 years ago

What volume of a 0.160 MM solution of KOHKOH must be added to 550.0 mLmL of the acidic solution to completely neutralize all of

puteri [66]

The given question is incomplete. The complete question is :

What volume of 0.160 M solution of KOH must be added to 550.0 mL of the acidic solution to completely neutralize all of the 0.150 M hydrochloric acid?

Answer: Volume in liters to three significant figures is 0.516 L

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH

We are given:

$n_1=1\\M_1=0.50M\\V_1=550.0mL\\n_2=1\\M_2=0.160M\\V_2=?$

Putting values in above equation, we get:

$1\times 0.150\times 550.0=1\times 0.160\times V_2\\\\V_2=516mL=0.516L$ (1L=1000ml)

Thus volume in liters to three significant figures is 0.516 L

6 0

3 years ago

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student genera

Paraphin [41]

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3

3 0

3 years ago