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GarryVolchara [31]
3 years ago
14

I need the correct answer

Chemistry
2 answers:
zalisa [80]3 years ago
6 0

Answer:

d.

Explanation:

They can be spread through contaminated surfaces.

natulia [17]3 years ago
3 0

Answer:

d

Explanation:  pathogens csnnot be spread be sontaminated surfesec

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What is the percent error for the experiment if the actual density is
OLga [1]

Answer:

<h3>The answer is 40.96%</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 2.49g/mL

error = 2.49 - 1.47 = 1.02

We have

p(\%) =  \frac{1.02}{2.49}  \times 100 \\ = 40.96385542...

We have the final answer as

<h3>40.96 %</h3>

Hope this helps you

4 0
3 years ago
Read 2 more answers
Identify the acid, base, conjugate acid and conjugate base in the following reactions:
Leni [432]

Answer:

a. NH₃ : base

CH₃COOH (acetic acid) : acid

NH₄⁺ : conjugate acid

CH₃COO⁻ : conjugate base

b. HClO₄ (perchloric acid) : acid

NH₃ : base

ClO₄⁻ : conjugate base

NH₄⁺ : conjugate acid

Hope this helps.

4 0
3 years ago
A gas occupies 800ml at a temperature of 27C. What is the volume at 132C?
pishuonlain [190]

V
1
​
/T
1
​
=V
2
​
/T
2
​

(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.

Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
3 0
3 years ago
In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit
Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone  phosphate (DHAP).  In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.  

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0
3 years ago
How many moles are in 1.05 g of gold (Au)?
Wittaler [7]

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

<u>Hence , moles can be calculated as -</u>

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

7 0
3 years ago
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