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3 years ago

Temperature and silica content determine the _______ of magma.

1 answer:

7 0

I think the correct answer is that they can determine the viscosity of the magma. The viscosity of a magma is largely controlled by the temperature, composition and the gas content. Also, silica content can define a magma type. It is said that higher silica content magma has a higher viscosity than those with lower silica content.

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How many milliliters of a 1.25 molar hydrochloric acid (HCl) solution would be needed to react completely with 60.0 grams of cal

stich3 [128]

Answer:

2400 mL

Explanation:

$Ca + 2HCl \implies CaCl_2 + H_2$

According to this equation, the stoichiometric ratio between $Ca$ and $HCl$ for the complete reaction is 1:2.

We know that the number of moles of $Ca$ can be calculated using the mole formula. (number of moles = mass / molar mass)

Moles of Calcium = $\frac{60}{40}$ = 1.5 mol

So the moles of $HCl$ = $1.5 \times 2$ = 3.0 mol

Volume of HCl solution = Moles of HCl/ concentration of HCl

Volume of HCl solution = $\frac{3}{1.25}$ = 2400 mL

4 0

3 years ago

4) The initial rate of the reaction between substances P and Q was measured in a series of

ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

$rate = k[P]^{2} [Q]$

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

$rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }$

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

$k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4$

Second row:

2. Rate value:

$rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1$

3.Third row:

$[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}$

4. Fourth row:

$[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}$

6 0

2 years ago

How many formula units of sodium chloride (NaCl) may theoretically be produced from 13.0 g FeCl3?

algol [13]

Answer:

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

Explanation:

Mass of ferric chloride = 13.0 g

Moles of ferric chloride = $\frac{13.0 g}{162.5 g/mol}=0.08 mol$

1 mole of ferric chloride has three moles of chloride ions.Then 0.08 moles of ferric chloride will have :

$3\times 0.08 mol=0.24 mol$ of chloride

$Na^++Cl^-\rightarrow NaCl$

1 mole of sodium ion reacts with 1 mole of chloride ion to form 1 mole of NaCl. Then 0.24 moles of chloride ion will give:

$\frac{1}{1}\times 0.24 mol=0.24 mol$ of NaCl

1 mole = $N_A=6.022\times 10^{23}$ molecules/ atoms

Number of NaCl molecules in 0.24 moles :

$=6.022\times 10^{23}\times 0.24=1.445\times 10^{23} molecules$

$1.445\times 10^{23}$ formula units of sodium chloride can be formed from 13.0 gram of ferric chloride.

4 0

3 years ago

In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)

Veronika [31]

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

4 0

2 years ago

Which one is idk what it is giving brainliest

Salsk061 [2.6K]

Answer:

B or C

Explanation:

I don't really know sorry.

8 0

2 years ago

Read 2 more answers