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Aleksandr [31]
3 years ago
14

Please help me, quick No links !

Chemistry
1 answer:
Simora [160]3 years ago
6 0

"One possibility is that the battle between the virus and your immune system can take as long as two weeks.

It could be the immune system holds the virus at bay,” said Tompkins.

Or, your immune system has to work so hard that after two weeks it’s inflamed and that’s what makes you feel bad.

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NaCl + H2SO4 ---&gt; Na2SO4 + HCl<br> Balance the double replacement chemical reaction.
VikaD [51]

Answer:

2NaCl+H2SO4-->Na2SO4+2HCl

Explanation:

There are two Na on the right, so put a 2 in front of NaCl on the left. This makes 3 Cl also, so put a 2 in front of HCl on the right. There are already 2 H on the left, so the equation is balanced.

7 0
4 years ago
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irina [24]

Answer:

H. The ermine's dark brown coat  in the summer changes to white is the winter.

Explanation:

If the ermine want to survive and not get attacked by predators, then they need to change the color of the coat from brown to white

4 0
3 years ago
Based on his education what is amazing about issac Newton’s great abilities in math and science
deff fn [24]

Answer:

Calculus.... Differentiation and integration..... This two... is mathematical application... was the key of revealing the hidden mysteries of world

7 0
3 years ago
Please write the answer in format below
jeyben [28]

Answer:

0.95 L

Explanation:

Step 1: Given data

Concentration of the Mg(NO₃)₂ solution (C): 0.32 M (0.32 mol/L)

Mass of Mg(NO₃)₂ (solute): 45 g

Step 2: Calculate the moles corresponding to 45 g of Mg(NO₃)₂

The molar mass of Mg(NO₃)₂ is 148.33 g/mol.

45 g Mg(NO₃)₂ × 1 mol Mg(NO₃)₂ /148.33 g Mg(NO₃)₂ = 0.303 mol Mg(NO₃)₂

Step 3: Calculate the volume of solution that contains 0.303 moles of Mg(NO₃)₂

The concentration of the solution is 0.32 M, that is, there are 0.32 moles of Mg(NO₃)₂ per liter of solution.

0.303 mol Mg(NO₃)₂ × 1 L Solution / 0.32 mol Mg(NO₃)₂ = 0.95 L

4 0
3 years ago
When 108 g of water at a temperature of 23.9 °c is mixed with 66.9 g of water at an unknown temperature, the final temperature o
olganol [36]

Here,

Heat gain by the first sample of water + Heat lost by the second sample of water is equal to zero (0).

Now, Mass of water sample one = 108 g (given)

Mass of water sample two = 66.9 g (given)

Temperature for water sample one = 23.9^{0}C

Let, temperature for water sample two =x

And, final temperature = 47.2^{0}C

Now,

mass of water sample one\times specific heat of water\times (T_{f} - T_{i}) + mass water sample two\times specific heat of water\times (T_{f} - T_{i}) = 0

where, T_{f} = final temperature

T_{i} = initial temperature

Substitute all the given values in above formula:

(108 g\times 4.184 J/g . K\times ( 47.2^{0}C- 23.9^{0}C) )+ (66.9 g \times 4.184 J/g . K\times (47.2^{0}C -x))= 0

(451.872 J/K \times (23.3^{0}C)) + (279.9096 J/K\times (47.2^{0}C -x)) = 0

(10528.6176 J/K(^{0}C)+ (13211.73312 J/K(^{0}C) -279.9096 J/K\times x)= 0

(23740.35072 J/K(^{0}C) -279.9096 J/K\times x)= 0

-279.9096 J/K\times x= -23740.35072 J/K(^{0}C)

x =\frac{23740.35072 J/K(^{0}C)}{279.9096 J/K}

[tex x =84.81^{0}C [/tex]









5 0
4 years ago
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